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DearCs  Stereotype  Edition, 

ELEMENTS 

OF 

GEOMETRY: 

CONTAINING 

THE    FIRST    SIX    BOOKS    OF    EUC 

WITH  A 

SUPPLEMENT       / 

ON  THE 

QUADRATURE  OF  THE  CIRCLE,  AND  THE 
GEOMETRY  OF  SOLIDS : 

TO  WHICH  ARE  ADDED, 

ELEMENTS  OF  PLANE  AND  SPHERICAL 
TRIGONOMETRY. 


BY 

JOHN  PLAYFAIR,  F.R.S.  Lond.  &  Edin. 

PBOrSSSOB  OF  NXTUBAL  PHILOSOPHY,  FOHMERLT  OF  MATHIMATICS,  IN  THE 
VNXVEESITY  OF  EDINBUBOH. 


FROM  THE  LAST  LONDON  EDITION,  ENLARGED. 


NEW  YORK: 
W.  E.  DEAN,  PRINTER,  &  PUBLISHER. 

1846. 


Entered  according  to  the  Act  of  Congress^  in  the  year  One  Thousand 
Eight  Hundred  and  Forty-five^  hy  W.  E.  Dean,  in  the  Clerk's  Of- 
fice  of  the  Southern  District  of  New-York, 


03 


PREFACE. 


It  is  a  remarkable  fact  in  the  history  of  science,  that  the  oldest  book  of 
Elementary  Geometry  is  still  considered  as  the  best,  and  that  the  writings 
of  Euclid,  at  the  distance  of  two  thousand  years,  continue  to  form  the  most 
approved  introduction  to  the  mathematical  sciences.  This  remarkable 
distinction  the  Greek  Geometer  owes  not  only  to  the  elegance  and  correct- 
ness of  his  demonstrations,  but  to  an  arrangement  most  happily  contrived 
for  the  purpose  of  instruction, — advantages  which,  when  they  reach  a  cer- 
tain eminence,  secure  the  works  of  an  author  against  the  injuries  of  time 
more  effectually  than  even  originality  of  invention.  The  Elements  of  Eu- 
clid, however,  in  passing  through  the  hands  of  the  ancient  editors  during 
the  decline  of  science,  had  suffered  some  diminution  of  their  excellentie,  and 
much  skill  and  learning  have  been  employed  by  the  modern  mathemati- 
cians to  deliver  them  from  blemishes  which  certainly  did  not  enter  into  their 
original  composition.  Of  these  mathematicians,  Dr.  Simson,  as  he  may 
be  accounted  the  last,  has  also  been  the  most  successful,  and  has  left  very 
little  room  for  the  ingenuity  of  future  editors  to  be  exercised  in,  either  by 
amending  the  text  of  Euclid,  or  by  improving  the  translations  from  it. 

Such  being  the  merits  of  Dr.  Simson's  edition,  and  the  reception  it  has 
met  with  having  been  every  way  suitable,  the  work  now  offered  to  the  pub- 
lic will  perhaps  appear  unnecessary.  And  indeed,  if  the  geometer  just 
named  had  written  with  a  view  of  accommodating  the  Elements  of  Euclid 
to  the  present  state  of  the  mathematical  sciences,  it  is  not  likely  that  any 
thing  new  in  Elementaiy  Geometry  would  have  been  soon  attempted.  But 
his  design  was  different;  it  was  his  object  to  restore  the  writings  of  Euclid 
to  their  original  perfection,  and  to  give  them  to  Modern  Europe  as  nearly 
as  possible  in  the  state  wherein  they  made  their  first  appearance  in  Ancient 
Greece.  For  this  undertaking,  nobody  could  be  better  qualified  than  Dr. 
SiMSON  ;  who,  to  an  accurate  knowledge  of  the  learned  languages,  and  an 
indefatigable  spirit  of  research,  added  a  profound  skill  in  the  ancient  Geome- 
try, and  an  admiration  of  it  almost  enthusiastic.  Accordingly,  he  not  only 
restored  the  text  of  Euclid  wherever  it  had  been  corrupted,  but  in  some 
cases  removed  imperfections  that  probably  belonged  to  the  original  work  : 
though  his  extreme  partiality  for  his  author  never  permitted  him  to  suppose 
that  such  honour  could  fall  to  the  share  either  of  himself,  or  of  any  other  of 
the  moderns. 

But,  after  all  this  was  accomplished,  something  still  remained  to  be  done, 
since,  notwithstanding  the  acknowledged  excellence  of  Euclid's  Ele- 
ments, it  could  not  be  doubted  that  some  alterations  might  be  made  that 
would  accommodate  them  better  to  a  state  of  the  mathematical  sciences,  so 
much  more  improved  and  extended  than  at  the  period  when  they  were 
written.  Accordingly,  the.object  of  the  edition  now  offered  to  the  public, is 
not  so  much  to  give  the  writings  of  Euclid  the  form  which  they  originally 
had,  as  that  which  may  at  present  render  them  most  useful. 


jib':' 


PREFACE. 

One  ol  the  alterations  made  with  this  view,  respects  the  Doctrine  of 
Proportion,  the  method  of  treating  which,  as  it  is  laid  down  in  the  fifth  of 
Euclid,  has  great  advantages  accompanied  with  considerable  defects  ;  of 
which,  however,  it  must  be  observed,  that  the  advantages  are  essential,  and 
the  defects  only  accidental.  To  explain  the  nature  of  the  former  requires 
a  more  minute  examination  than  is  suited  to  this  place,  and  must  therefore 
be  reserved  for  the  Notes  ;  but,  in  the  mean  time,  it  may  be  remarked,  that 
no  definition,  except  that  of  Euclid,  has  ever  been  given,  from  which  the 
properties  of  proportionals  can  be  deduced  by  reasonings,  which,  at  the 
same  time  that  they  are  perfectly  rigorous,  are  also  simple  and  direct.  As 
to  the  defects,  the  prolixness  and  obscurity  that  have  so  often  been  com- 
plained of  in  the  fifth  Book,  they  seem  to  arise  chiefly  from  the  nature  of 
the  language  employed,  which  being  no  other  than  that  of  ordinary  dis- 
course, cannot  express,  without  much  tediousness  and  circumlocution,  the 
relations  of  mathematical  quantities,  when  taken  in  their  utmost  generality, 
and  when  no  assistance  can  be  received  from  diagrams.  As  it  is  plain  that 
the  concise  language  of  Algebra  is  directly  calculated  to  remedy  this  in- 
convenience, I  have  endeavoured  to  introduce  it  here,  in  a  very  simple  form 
however,  and  without  changing  the  nature  of  the  reasoning,  or  departing 
in  any  thing  from  the  rigour  of  geometrical  demonstration.  By  this  means, 
the  steps  of  the  reasoning  which  were  before  far  separated,  are  brought 
near  to  one  another,  and  the  force  of  the  whole  is  so  clearly  and  directly 
perceived,  that  I  am  persuaded  no  more  difiiculty  will  be  found  in  under- 
standing the  propositions  of  the  fifth  Book  than  those  of  any  other  of  the 
Elements. 

In  the  second  Book,  also,  some  algebraic  signs  have  been  introduced,  for 
the  sake  of  representing  more  readily  the  addition  and  subtraction  of  the 
rectangles  on  which  the  demonstrations  depend.  The  use  of  such  sym- 
bolical writing,  in  translating  from  an  original,  where  no  symbols  are  used, 
cannot,  I  think,  be  regarded  as  an  unwarrantable  liberty  :  for,  if  by  that 
means  the  translation  is  not  made  into  English,  it  is  made  into  that  univer- 
sal language  so  much  sought  after  in  all  the  sciences,  but  destined,  it  would 
seem,  to  be  enjoyed  only  by-the  mathematical. 

The  alterations  above  mentioned  are  the  most  material  that  have  been 
attempted  on  the  books  of  Euclid.  There  are,  however,  a  few  others, 
which,  though  less  considerable,  it  is  hoped  may  in  some  degree  facilitate 
the  study  of  the  Elements.  Such  are  those  made  on  the  definitions  in  the 
first  Book,  and  particularly  on  that  of  a  straight  line.  A  new  axiom  is  also 
introduced  in  the  room  of  the  12th,  for  the  purpose  of  demonstrating  more 
easily  some  of  the  properties  of  parallel  lines.  In  the  third  Book,  the  re- 
marks concerning  the  angles  made  by  a  straight  line,  and  the  circumference 
of  a  circle,  are  left  out,  as  tending  to  perplex  one  who  has  advanced  no 
farther  than  the  elements  of  the  science.  Some  propositions  also  have 
been  added  ;  but  for  a  fuller  detail  concerning  these  changes,  I  must  refer 
to  the  Notes,  in  which  several  of  the  more  difficult,  or  more  interesting  sub- 
jects of  Elementary  Geometry  are  treated  at  considerable  length. 

College  of  Edinburgh, 
Dec.  1,  1813. 


ELEMEMTS 


OF 


GEOMETRY. 


BOOK  I. 

THE  PRINCIPLES. 

EXPLANATION  OF  TERMS  AND  SIGNS. 

1.  Geometry  is  a  science  which  has  for  its  object  the  measurement  of  mag- 
nitudes. 

Magnitudes  may  be  considered  under  three  dimensions, — length,  breadth, 
height  or  thickness. 

2.  In  Geometry  there  are  several  general  terms  or  principles ;  such  as, 
Definitions,  Propositions,  Axioms,  Theorems,  Problems,  Lemmas,  Scho- 
liums, Corollaries,  &c. 

3.  A  Definition  is  the  explication  of  any  term  or  word  in  a  science,  show- 
ing the  sense  and  meaning  in  which  the  term  is  employed. 

Every  definition  ought  to  be  clear,  and  expressed  in  words  that  are 
common  and  perfectly  well  understood. 

4.  An  Axiom,  or  Maxim,  is  a  self-evident  proposition,  requiring  no  formal 
demonstration  to  prove  the  truth  of  it ;  but  is- received  and  assented  to  as 
soon  as  mentioned. 

Such  as,  the  whole  of  any  thing  is  greater  than  a  part  of  it ;  or,  the 
whole  is  equal  to  all  its  parts  taken  together ;  or,  two  quantities  that 
are  each  of  them  equal  to  a  third  quantity,  are  equal  to  each  other. 

5.  A  Theorem  is  a  demonstrative  proposition  ;  in  which  some  property  is 
asserted,  and  the  truth  of  it  required  to  be  proved. 

Thus,  when  it  is  said  that  the  sum  of  the  three  angles  of  any  plane  tri- 
angle is  equal  to  two  right  angles,  this  is  called  a  Theorem;  and  the 
method  of  collecting  the  several  arguments  and  proofs,  and  laying 
them  together  in  proper  order,  by  means  of  which  the  truth  of  the 
proposition  becomes  evident,  is  called  a  Demonstration. 

6.  A  Direct  Demonstration  is  that  which  concludes  with  the  direct  and  cer- 
tain proof  of  the  proposition  in  hand. 

It  is  also  called  Positive  or  Affirmative,  and  sometimes  an  Ostensive  De- 
monstration,  because  it  is  most  satisfactory  to  the  mind. 


I 


6  ELEMENTS 

7.  An  Indirect  or  Negative  Demonstration  is  that  which  shows  a  proposition 
to  be  true,  by  proving  that  some  absurdity  would  necessarily  follow  if 
the  proposition  advanced  were  false. 

This  is  sometimes  called  Reductio  ad  Ahsurdum ;  because  it  shows  the 
absurdity  and  falsehood  of  all  suppositions  contrary  to  that  contained 
in  the  proposition. 

8.  A  Problem  is  a  proposition  or  a  question  proposed,  which  requires  a  so- 
lution. 

As,  to  draw  one  line  perpendicular  to  another  ;  or  to  divide  a  line  into 
two  equal  parts. 

9.  Solution  of  a  problem  is  the  resolution  or  answer  given  to  it. 

A  Numerical  or  Numeral  solution,  is  the  answer  given  in  numbers.  A 
Geometrical  solution,  is  the  answer  given  by  the  principles  of  Geome- 
try.    And  a  Mechanical  solution,  is  one  obtained  by  trials. 

10.  A  Lemma  is  a  preparatory  proposition,  laid  down  in  order  to  shorten 
the  demonstration  of  the  main  proposition  which  follows  it. 

1 1 .  A  Corollary^  or  Consectary,  is  a  consequence  drawn  immediately  from 
some  proposition  or  other  premises. 

12.  A  Scholium  is  a  remark  or  observation  made  on  some  foregoing  propo- 
sition or  premises. 

13.  An  Hypothesis  is  a  supposition  assumed  to  be  true,  in  order  to  argue 
from,  or  to  found  upon  it  the  reasoning  and  demonstration  of  some  pro- 
position. 

14.  A  Postulate,  or  Petition,  is  something  required  to  be  done,  which  is  so 
easy  and  evident  that  no  person  will  hesitate  to  allow  it. 

15.  Method  is  the  art  of  disposing  a  train  of  arguments  in  a  proper  order, 
to  investigate  the  truth  or  falsity  of  a  proposition,  or  to  demonstrate  it  to 
others  when  it  has  been  found  out.  This  is  either  Analytical  or  Syn- 
thetical. 

16.  Analysis,  or  the  Analytic  method,  is  the  art  or  mode  of  finding  out  the 
truth  of  a  proposition,  by  first  supposing  the  thing  to  be  done,  and  then 
reasoning  step  by  step,  till  we  arrive  at  some  known  truth.  This  is  also 
called  the  Method  of  Invention,  or  Resolution ;  and  is  that  which  is  com* 
monly  used  in  Algebra. 

17.  Synthesis,  or  the  Synthetic  Method,  is  the  searching  out  truth,  by  first 
laying  down  simple  principles,  and  pursuing  the  consequences  flowing 
from  them  till  we  arrive  at  the  conclusion.  This  is  also  called  the  Me- 
thod of  Composition ;  and  is  that  which  is  commonly  used  in  Geometry. 

18.  The  sign  =  (or  two  parallel  lines),  is  the  sign  of  equality ;  thus, 
A=B,  implies  that  the  quantity  denoted  by  A  is  equal  to  the  quantity 
denoted  by  B,  and  is  read  A  equal  to  B. 

19.  To  signify  that  A  is  greater  than  B,  the  expression  A  7  B  is  used.  And 
to  signify  that  A  is  less  than  B,  the  expression  A/  B  is  used. 


OF  GEOMETRY.     BOOK  I.  7 

20.  The  sign  of  Addition  is  an  erect  cross  ;  thus  A+B  implies  the  sum  of 
A  and  B,  and  is  called  A  plus  B. 

21.  Subtraction  is  denoted  by  a  single  line  ;  as  A — B,  which  is  read  A 
minus  B  ;  A — B  represents  their  difference,  or  the  part  of  A  remaining, 
when  a  part  equal  to  B  has  been  taken  away  from  it. 

In  like  manner,  A — B+C,  or  A+C — B,  signifies  that  A  and  0  are  to 
be  added  together,  and  that  B  is  to  be  subtracted  from  their  sum. 

22.  Multiplication  is  expressed  by  an  oblique  cross,  by  a  point,  or  by  simple 
apposition:  thus,  AxB,  A  .  B,  or  AB,  signifies  that  the  quantity  de- 
noted by  A  is  to  be  multiplied  by  the  quantity  denoted  by  B.  The  ex- 
pression AB  should  not  be  employed  when  there  is  any  danger  of  con- 
founding it  with  that  of  the  line  AB,  the  distance  between  the  points  A 
and  B.  The  multiplication  of  numbers  cannot  be  expressed  by  simple 
apposition. 

23.  When  any  quantities  are  enclosed  in  a  parenthesis,  or  have  aline  drawn 
over  them,  they  are  considered  as  one  quantity  with  respect  to  other 
symbols:  thus,  the  expression  AX(B+C — D),  or  AxB-|-C — D,  re- 
presents the  product  of  A  by  the  quantity  B-fC — D.  In  like  manner, 
(A+B)X(A — B  4-0),  indicates  the  product  of  A+B  by  the  quantity 
A— B+C. 

24.  The  Co-efficient  of  a  quantity  is  the  number  prefixed  to  it:  thus,  2AB 
signifies  that  the  line  AB  is  to  be  taken  2  times ;  JAB  signifies  the  half 
of  the  line  AB. 

25.  Division^  oi  the  ratio  of  one  quantity  to  another,  is  usually  denoted  by 
placing  one  of  the  two  quantities  over  the  other,  in  the  formof  a  fraction : 

thus,  —  signifies  the  ratio  or  quotient  arising  from  the  division  of  the 

quantity  A  by  B.     In  fact,  this  is  division  indicated. 

26.  The  SquarCj  Cuhcj  &c.  of  a  quantity,  are  expressed  by  placing  a  small 
figure  at  the  right  hand  of  the  quantity:  thus,  the  square  of  the  line 
AB  is  denoted  by  AB^,  the  cube  of  the  line  AB  is  designated  by  AB'^ ; 
and  so  on. 

27.  The  Roots  of  quantities  are  expressed  by  means  of  the  radical  sign  ■/, 
with  the  proper  index  annexed ;  thus,  the  square  root  of  5  is  indicated 

■v/5  ;  y(A.  X  B)  means  the  square  root  of  the  product  of  A  and  B,  or  the 
mean  proportional  between  them.  The  roots  of  quantities  are  some- 
times expressed  by  means  of  fractional  indices  :  thus,  the  cube  root  of 

A  X  B  X  C  may  be  expressed  by  V^^^^xC,  or  (A  X  B  X  Cp,  and 
so  on. 

28.  Numbers  in  a  parenthesis,  such  as  (15.  1.),  refers  back  to  the  number 
of  the  proposition  and  the  Book  ia  which  it  has  been  announced  or  de- 
monstrated. The  expression  (15.  1.)  denotes  the  fifteenth  proposition, 
first  book,  and  so  on.  In  like  manner,  (3.  Ax.)  designates  the  third 
axiom ;  (2.  Post.)  the  second  postulate ;  (Def.  3.)  the  third  definition, 
and  so  on. 


8  ELEMENTS 

29.  The  word,  therefore^  or  hence,  frequently  occurs.     To  express  eitlier  of 
these  words,  the  sign  .*.  is  generally  used. 

30.  If  the  quotients  of  two  pairs  of  numbers,  6r  quantities,  are  equal,  the 

quantities  are  said  to  he  proportional:  thus,  if  —  =  jT  ;  then,  A  is  to  B 

as  C  to  D.     And  the  abbreviations  of  th€  proportion  is,  A  :  B  : :  C  :  D ; 
it  is  sometimes  written  A  :  B=C  :  D. 


DEFINITIONS. 

1.'  "A  Point  is  that  which  has  position,  but  not  magnitude*."  (See 
Notes.) 

2.  A  line  is  length  without  breadth. 

"  Corollary.  The  extremities  of  a  line  are  points  ;  and  the  intersections 
"  of  one  line  with  another  are  also  points." 

3.  "  If  two  lines  are  such  that  they  cannot  coincide  in  any  two  points,  with- 
"  out  coinciding  altogether,  each  of  them  is  called  a  straight  line." 

"  Cor.  Hence  two  straight  lines  cannot  inclose  a  space.  Neither  can  two 
"  straight  lines  have  a  common  segment ;  that  is,  they  cannot  coincide 
"  in  part,  without  coinciding  altogether." 

4.  A  superficies  is  that  which  has  only  length  and  breadth. 

*  CoR.  The  extremities  of  a  superficies  are  lines  ;  and  the  intersections  of 
"  one  superficies  with  another  are  also  lines." 

5.  A  plane  superficies  is  that  in  which  any  two  points  being  taken,  the 
straight  line  between  them  lies  wholly  in  that  superficies. 

6.  A  plane  rectilineal  angle  is  the  inclination  of  two  straight  lines  to  one 
another,  which  meet  together,  but  are  not  in  the  same  straight  line. 


B 


N.  B.  'When  several  angles  are  at  one  point  B,  any  one  of  them  is  ex- 

*  pressed  by  three  letters,  of  which  the  letter  that  is  at  the  vertex  of  the  an- 

*  gle,  that  is,  at  the  point  in  which  the  straight  lines  that  contain  the  angle 

*  meet  one  another,  is  put  between  the  other  two  letters,  and  one  of  these 

*  two  is  somewhere  upon  one  of  those  straight  lines,  and  the  other  upon  the 

*  other  line :  Thus  the  angle  which  is  contained  by  the  straight  lines,  AB, 

*  CB,  is  named  the  angle  ABC,  or  CBA  ;  that  which  is  contained  by  AB, 

*  The  definitions  marked  with  inverted  commas  are  different  from  those  of  Kuclid. 


OF  GEOMETRY.     BOOK  I. 


d 


*  BD,  is  named  the  angle  ABD,  or  DBA  ;  and  that  which  is  contained  by 

*  BD,  CB,  is  called  the  angle  DBC,  or  CBD  ;  but,  if  there  be  only  one  an- 
'  gle  at  a  point,  it  may  be  expressed  by  a  letter  placed  at  that  point ;  as  the 
'  angle  at  E.' 


7.  When  a  straight  line  standing  oh  another 
straight  line  makes  the  adjacent  angles  equal 
to  one  another,  each  of  the  angles  is  called 
a  right  angle  ;  and  the  straight  line  which 
stands  on  the  other,  is  called  a  perpendicu- 
lar to  it. 


8.  An  obtuse  angle  is  that  which  is  greater  than  a  right  angle. 


9.  An  acute  angle  is  that  which  is  less  than  a  right  angle. 

10.  A  figure  is  that  which  is  enclosed  by  one  or  more  boundaries. — The 
word  area  denotes  the  quantity  of  space  contained  in  a  figure,  without  any 
reference  to  the  nature  of  the  line  or  lines  which  bound  it, 

11.  A  circle  is  a  plane  figure  contained  by  one  line,  which  is  called  the 
circumference,  and  is  such  that  all  straight  lines  drawn  from  a  certain 
point  within  the  figure  to  the  circumference,  are  equal  to  one  another. 


12.  And  this  point  is  called  the  centre  of  the  circle. 

13.  A  diameter  of  a  circle  is  a  straight  line  drawn  through  the  centre,  and 
terminated  both  ways  by  the  circumference. 

14.  A  semicircle  is  the  figure  contained  by  a  diameter  and  the  part  of  the 
circumference  cut  off  by  the  diameter, 

2 


10 


ELEMENTS 


15.  Rectilineal  figures  are  those  which  are  contained  by  straight  lines. 

16.  Trilateral  figures,  or  triangles,  by  three  straight  lines. 

17.  Quadrilateral,  by  four  straight  lines. 

18.  Multilateral  figures,  or  polygons,  by  more  than  four  straight  lines. 

19.  Of  three  sided  figures,  an  equilateral  triangle  is  that  which  has  three 
equal  sides. 

20.  An  isosceles  triangle  is  that  which  has  only  two  sides  equal. 


21.  A  scalene  triangle  is  that  which  has  three  unequal  sides. 

22.  A  right  angled  triangle  is  that  which  has  a  right  angle. 

23.  An  obtuse  angled  triangle  is  that  which  has  an  obtuse  angle. 


24.  An  acute  angled  triangle  is  that  which  has  three  acute  angles. 

25.  Of  four  sided  figures,  a  square  is  that  which  has  all  its  sides  equal 
and  all  its  angles  right  angles. 


26.  An  oblong  is  that  which  has  all  its  angles  right  angles,  but  has  not  all 
its  sides  equal. 

27.  A  rhombus  is  that  which  has  all  its  sides  equal,  but  its  angles  are  not 
right  angles. 


OF  GEOMETRY.    BOOK  I.  U 

28.  A  rhomboid  is  that  which  has  its  opposite  sides  equal  to  one  another, 
but  all  its  sides  are  not  equal,  nor  its  angles  right  angles. 

29.  All  other  four  sided  figures  besides  these,  are  called  trapeziums. 

30.  Parallel  straight  lines  are  such  as  are  in  the  same  plane,  and  which, 
being  produced  ever  so  far  both  ways,  do  not  meet. 


POSTULATES. 

1.  Let  it  be  granted  that  a  straight  line  may  be  drawn  from  any  one  point 
td  any  other  point. 

2.  That  a  terminated  straight  line  may  be  produced  to  any  length  in  a 
straight  line. 

3.  And  that  a  circle  may  be  described  from  any  centre,  at  any  distance 
from  that  centre. 


AXIOMS. 

1 .  Things  which  are  equal  to  the  same  thing  are  equal  to  one  another. 

2.  If  equals  be  added  to  equals,  the  wholes  are  equal. 

3.  If  equals  be  taken  from  equals,  the  remainders  are  equal. 

4.  If  equals  be  added  to  unequals,  the  wholes  are  unequal. 

5.  If  equals  be  taken  from  unequals,  the  remainders  are  unequal. 

6.  Things  which  are  doubles  of  the  same  thing,  are  equal  to  one  another. 

7.  Things  which  are  halves  of  the  same  thing,  are  equal  to  one  another. 

8.  Magnitudes  which  coincide  with  one  another,  that  is,  which  exactly 
fill  the  same  space,  are  equal  to  one  another. 

9.  The  whole  is  greater  than  its  part. 

10.  All  right  angles  are  equal  to  one  another. 

11."  Two  straight  lines  which  intersect  one  another,  cannot  be  both  pa- 
"  rallel  to  the  same  straight  line." 


12 


ELEMENTS 


PROPOSITION  I.     PROBLEM.  '■  ^^^ 

To  describe  an  equilateral  triangle  upon  a  given  finite  straight  line. 

Let  AB  be  the  given  straight  line  ;  it  is  required  to  describe  an  equi- 
lateral triangle  upon  it. 

From  the  centre  A,  at  the  dis- 
tance AB,  describe  (3.  Postulate) 
the  circle  BCD,  and  from  the  cen- 
tre B,  at  the  distance  BA,  describe 
the  circle  ACE  ;  and  from  the  point 
C,  in  which  the  circles  cut  one  an- 
other, draw  the  straight  lines  (1. 
Post.)  CA,  CB  to  the  points  A,  B  ; 
ABC  is  an  equilateral  triangle. 

Because  the  point  A  is  the  cen- 
tre of  the  circle  BCD,  AC  is  equal  if 
(11.  Definition)  to  AB  ;  and  because  the  point  B  is  the  centre  of  the  cir- 
cle ACE,  BC  is  equal  to  AB  :  But  it  has  been  proved  that  CA  is  equal 
to  AB ;  therefore  CA,  CB  are  each  of  them  equal  to  AB  ;  now  things 
which  are  equal  to  the  same  are  equal  to  one  another,  (1.  Axiom) ;  there- 
fore CA  is  equal  to  CB  ;  wherefore  CA,  AB,  CB  are  equal  to  one  another  ; 
and  the  triangle  ABC  is  therefore  equilateral,  and  it  is  described  upon  the 
given  straight  line  AB. 

PROP.  II.  ■  PROB. 

From  a  given  point  to  draw  a  straight  line  equal  to  a  given  straight  line. 

Let  A  be  the  given  point,  and  BC  the  given  straight  line ;  it  is  required 
to  draw,  from  the  point  A,  a  straight  line  equal  to  BC. 

From  the  point  A  to  B  draw  (1.  Post.) 
the  straight  line  AB  ;  and  upon  it  describe 
(1.  1.)  the  equilateral  triangle  DAB,  and 
produce  (2.  Post.)  the  straight  lines  DA, 
BD,  to  E  and  F  ;  from  the  centre  B,  at  the 
distance  BC,  describe  (3.  Post.)  the  circle 
CGH,  and  from  the  centre  D,  at  the  dis- 
tance DG,  describe  the  circle  GKL,  AL  is 
equal  to  BC. 

Because  the  point  B  is  the  centre  of  the 
circle  CGH,  BC  is  equal  (11.  Def.)  to  BG ; 
and  because  D  is  the  centre  of  the  circle 
GKL,  DL  is  equal  to  DG,  and  DA,  DB, 
parts  of  them,  are  equal ;  therefore  the  re- 
mainder AL  is  equal  to  .the  remainder  (3. 
Ax.)  BG:  But  it  has  been  shewn  that  BC  is  equal  to  BG  ;  wherefore 
AL  and  BC  are  each  of  them  equal  to  BG ;  and  things  that  are  equal 


OF  GEOMETRY.    BOOK  I. 


13 


to  the  same  are  equal'to  one  another  ;  therefore  the  straight  line  AL  is 
equal  to  BC.  Wherefore,  from  the  given  point  A,  a  straight  line  AL  has 
been  drawn  equal  to  the  given  straight  line  BC. 

PROP.  III.     PROB. 

From  the  greater  of  two  given  straight  lines  to  cut  off  a  part  equal  to  the 

less. 

Let  AB  and  C  be  the  two  given  straight 
lines,  whereof  AB  is  the  greater.  It  is 
required  to  cut  off  from  AB,  the  grteter, 
a  part  equal  to  C,  the  less. 

From  the  point  A  draw  (2.  1.)  the 
straight  line  AD  equal  to  C ;  and  from 
the  centre  A,  and  at  the  distance  AD, 
describe  (3.  Post.)  the  circle  DEF ;  and 
because  A  is  the  centre  of  the  circle 
DEF,  AE  is  equal  to  AD;  but  the 
straight  line  C  is  likewise  equal  to  AD  ; 

whence  AE  and  C  are  each  of  them  equal  to  AD  ;  wherefore  the  straight 
line  AE  is  equal  to  (1.  Ax.)  C,  and  from  AB  the  greater  of  two  straight 
lines,  a  part  AE  has  been  cut  off  equal  to  C  the  less. 

PROP.  IV.     THEOREM. 

If  two  triangles  have  two  sides  of  the  one  equal  to  two  sides  of  the  other,  each 
to  each ;  and  have  likewise  the  angles  contained  by  those  sides  equal  to 
one  another,  their  bases,  or  third  sides,  shall  be  equal ;  and  the  areas  of 
the  triangles  shall  be  equal ;  and  their  other  angles  shall  be  equal,  each  to 
each,  viz.  those  to  which  the  equal  sides  are  opposite* 

Let  ABC,  DEF  be  two  triangles  which  have  the  two  sides  AB,  AC 
equal  to  the  two  sides  DE,  DF,  each  to  each,  viz.  AB  to  DE,  and  AC  to 
DF ;  and  let  the  angle 
BAG  be  also  equal  to  the 
angle  EDF :  then  shall 
the  base  BC  be  equal  to 
the  base  EF  ;  and  the  tri- 
angle ABC  to  the  triangle 
DEF;  and  the  other  an- 
gles, to  which  the  equal 
sides  are  opposite,  shall  .^- 
be   equal,   each  to  each,         "  O  E  F 

riz.  the  angle  ABC  to  the  angle  DEF,  and  the  angle  ACB  to  DFE. 

For,  if  the  triangle  ABC  be  applied  to  the  triangle  DEF,  so  that  the 
point  A  may  be  on  D,  and  the  straight  line  AB  upon  DE  ;  the  point  B 
shall  coincide  with  the  point  E,  because  AB  is  equal  to  DE  ;  and  AB 


*  The  three  conclusions  in  this  enunciation  are  more  briefly  expressed  by  saying,  that  tht 
trhmglee  are  every  way  equal-. 


14  ELEMENTS 

coinciding  with  DE,  AC  shall  coincide  with  DF,  because  the  angle  BAG 
is  equal  to  the  angle  EDF  ;  wherefore  also  the  point  C  shall  coincide  with 
the  point  F,  because  AC  is  equal  to  DF  :  But  the  point  B  coincides  with 
the  point  E ;  wherefore  the  base  BC  shall  coincide  with  the  base  EF 
(cor.  def.  3.),  and  shall  be  equal  to  it.  Therefore  also  the  whole  triangle 
ABC  shall  coincide  with  the  whole  triangle  DEF,  so  that  the  spaces  which 
they  contain  or  their  areas  are  equal ;  and  the  remaining  angles  of  the 
one  shall  coincide  with  the  remaining  angles  of  the  other,  and  be  equal  to 
them,  viz.  the  angle  ABC  to  the  angle  DEF,  and  the  angle  ACB  to  the 
angle  DFE.  Therefore,  if  two  triangles  have  two  sides  of  the  one  equal 
to  two  sides  of  the  other,  each  to  each,  and  have  likewise  the  angles  con- 
tained by  those  sides  equal  to  one  another ;  their  bases  shall  be  equal, 
and  their  areas  shall  be  equal,  and  their  other  angles,  to  which  the  equal 
sides  are  opposite,  shall  be  equal,  each  to  each. 

PROP.  V.    THEOR. 

The  angles  at  the  base  of  an  Isosceles  triangle  are  equal  to  one  another  ;  and 
if  the  equal  sides  be  produced^  the  angles  upon  the  other  side  of  the  base 
shall  be  equal. 

Let  ABC  be  an  isosceles  triangle,  of  which  the  side  AB  is  equal  to  AC, 
and  let  the  straight  lines  AB,  AC  be  produced  to  D  and  E,  the  angle  ABC 
shall  be  equal  to  the  angle  ACB,  and  the  angle  CBD  to  the  angle  BCE. 

In  BD  take  any  point  F,  and  from  AE  the  greater  cut  off  AG  equal 
(3.  1.)  to  AF,  the  less,  and  join  FC,  GB. 

Because  AF  is  equal  to  AG,  and  AB  to  AC,  the  two  sides  FA,  AC  are  equal 
to  the  two  Gx\,  AB,  each  to  each ;  and  they  contain  the  angle  FAG  com- 
mon to  the  two  triangles,  AFC,  AGB ; 
therefore  the  base  FC  is  equal  (4.  1.)  to 
the  base  GB,  and  the  triangle  AFC  to 
the  triangle  AGB;  and  the  remaining 
angles  of  the  one  are  equal  (4.  1.)  to 
the  remaining  angles  of  the  other,  each  to 
each,  to  which  the  equal  sides  are  oppo- 
site, viz.  the  angte  ACF  to  the  angle 
ABG,  and  the  angle  AFC  to  the  angle 
AGB :  And  because  the  whole  AF  is 
equal  to  the  whole  AG,  and  the  part  AB 
to  the  part  AC  ;  the  remainder  BF  shall 
be  equal  (3.  Ax.)  to  the  remainder  CG ; 
and  FC  was  proved  to  be  equal  to  GB,      ^'  t-         \E 

therefore  the  two  sides  BF,  FC  are  equal  to  the  two  CG,  GB,  each  to 
each ;  but  the  angle  BFC  is  equal  to  the  angle  CGB  ;  wherefore  the  tri- 
angles BFC,  CGB  are  equal  (4. 1.),  and  their  remaining  angles  are  equal- 
to  which  the  equal  sides  are  opposite ;  therefore  the  angle  FBC  is  equal 
to  the  angle  GCB,  and  the  angle  BCF  to  the  angle  CBG.  Now,  since 
it  has  been  demonstrated,  that  the  whole  angle  ABG  is  equal  to  the  whole 
ACF,  and  the  part  CBG  to  the  part  BCF,  the  remaining  angle  ABC  is 
therefore  equal  to  the  remaining  angle  ACB,  which  are  die  angles  at  the 


OF  GEOMETRY.    BOOK  I.  15 

base  of  the  triangle  ABC :  And  it  has  also  been  proved  that  the  angle 
FBC  is  equal  to  the  angle  GCB,  which  are  the  angles  upon  the  other  side 
of  t?ie  base. 

Corollary.     Hence  every  equilateral  triangle  is  also  equiangular. 

PROP.  VI.    THEOR. 

If  two  angles  of  a  triangle  be  equal  to  one  another,  the  sides  which  subtend 
or  are  opposite  to  them,  are  also  equal  to  one  another. 

Let  ABC  be  a  triangle  having  the  angle  ABC  equal  to  the  angle  ACB  ; 
the  side  AB  is  also  equal  to  the  side  AC. 

For,  if  AB  be  not  equal  to  AC,  one  of  them  is 
greater  than  the  other :  Let  AB  be  the  greater, 
and  from  it  cut  (3.  1.)  off  DB  equal  to  AC  the 
less,  and  join  DC  ;  therefore,  because  in  the  tri- 
angles DBC,  ACB,  DB  is  equal  to  AC,  and  BC 
common  to  both,  the  two  sides  DB,  BC  are  equal 
to  the  two  AC,  CB,  each  to  each ;  but  the  angle 
DBC  is  also  equal  to  the  angle  ACB  ;  therefore 
the  base  DC  is  equal  to  the  base  AB,  and  the  area 
of  the  triangle  DBC  is  equal  to  that  of  the  triangle 
(4.  1.)  ACB,  the  less  to  the  greater  ;  which  is  ab- 
surd. Therefore,  AB  is  not  unequal  to  AC,  that 
is,  it  is  equal  to  it. 

CoR.     Hence  every  equiangular  triangle  is  also  equilateral. 

PROP.  VH.    THEOR. 

XJpon  the  same  base,  and  on  the  same  side  of  it,  there  cannot  be  two  triangles^ 
that  have  their  sides  which  are  terminated  in  one  extremity  of  the  base 
equal  to  one  another,  and  likewise  those  which  are  terminated  in  the  other 
extremity,  equal  to  one  another. 

Let  there  be  two  triangles  ACB,  ADB,  upon  the  same  base  AB,  and 
ujJon  the  same  side  of  it,  which  have  their  sides  CA,  DA,  terminated  in  A 
equal  to  one  another ;  then  their  sides  CB,  DB,  terminated  in  B,  cannot 
be  equal  to  one  another. 

Join  CD,  and  if  possible  let  CB  be  -^ 
equal  to  DB  ;  then,  in  the  case  in  which  ^ 
the  vertex  of  each  of  the  triangles  is  with- 
out the  other  triangle,  because  AC  is 
equal  to  AD,  the  angle  ACD  is  equal  (5. 
L)  to  the  angle  ADC  :  But  the  angle 
ACD  is  greater  than  the  angle  BCD  ; 
therefore  the  angle  ADC  is  greater  also 
than  BCD  ;  much  more  then  is  the  angle 
BDC  greater  than  the  angle  BCD.  Again, 
because  CB  is  equal  to  DB,  the  angle 
BDC  is  equal  (5.  1.)  to  the  angle  BCD ;      A.^ ^B 


16 


ELEMENTS 


but  it  has  been  demonstrated  to  be  greater  than  it ;  which  is  impossi 
ble. 

But  if  one  of  the  vertices,  as  D, 
be  within  the  other  triangle  ACB  ; 
produce  AC,  AD  to  E,  F ;  therefore, 
because  AC  is  equal  to  AD  in  the 
triangle  ACD,  the  angles  ECD,  FDC 
upon  the  other  side  of  the  base  CD 
are  equal  (5.  1.)  to  one  another,  but 
the  angle  ECD  is  greater  than  the 

angle  BCD  ;  wherefore   the  angle      .  „ 

FDC  is  likewise  greater  than  BCD  ;     ^  ^ 

much  more  then  is  the  angle  BDC  greater  than  the  angle  BCD.  Again, 
because  CB  is  equal  to  DB,  the  angle  BDC  is  equal  (5.  1.)  to  the  angle 
BCD ;  but  BDC  has  been  proved  to  be  greater  than  the  same  BCD ; 
which  is  impossible.  The  case  in  which  the  vertex  of  one  triangle  is 
upon  a  side  of  the  other,  needs  no  demonstration. 

Therefore,  upon  the  same  base,  and  on  the  same  side  of  it,  there  cannot 
be  two  triangles  that  have  their  sides  which  are  terminated  in  one  extrem- 
ity of  the  base  equal  to  one  another,  and  likewise  those  which  are  termina- 
ted in  the  other  extremity  equal  to  one  another. 

PROP.  VIII.    THEOR. 

If  two  triangles  have  two  sides  of  the  one  equal  to  two  sides  of  the  others 
each  to  each,  and  have  likewise  their  bases  equal ; '  the  angle  which  is  contain- 
ed by  the  two  sides  of  the  one  shall  be  equal  to  the  angle  contained  by  the  two 
sides  of  the  other. 

Let  ABC,  DEF  be  two  triangles  having  the  two  sides  AB,  AC,  equal 
to  the  two  sides  DE,  DF,  each  to  each,  viz.  AB  to  DE,  and  AC  to  DF ; 


and  also  the  base  BC  equal  to  the  base  EF.     The  angle  BAC.isn^qual  to 
the  angle  EDF. 

For,  if  the  triangle  ABC  be  applied  to  the  triangle  DEF,  so  that  the 
point  B  be  on  E,  and  the  straight  line  BC  upon  EF  ;  the  point  C  shall  also 
coincide  with  the  point  F,  because  BC  is  equal  to  EF  :  therefore  BC  coin- 
ciding with  EF,  BA  and  AC  shall  coincide  with  ED  and  DF  ;  for,  if 
BA  and  CA  do  not  coincide  with  ED  and  FD,  but  have  a  difierent  situa- 


OF  GEOMETRY.     BOOK  I. 


17 


tion,  as  EG  and  FG  ;  then,  upon  the  same  base  EF,  and  upon  the  same 
side  of  it,  there  can  be  two  triangles  EDF,  EGF,that  have  their  sides  which 
are  terminated  in  one  extremity  of  the  base  equal  to  one  another,  and  like- 
wise their  sides  terminated  in  the  other  extremity  ;  but  this  is  impossible 
(7.  1.) ;  therefore,  if  the  base  BC  coincides  with  the  base  EF,  the  sides 
BA,  AC  cannot  but  coincide  with  the  sides  ED,  DF ;  wherefore  likewise 
the  angle  BAG  coincides  with  the  angle  EDF,  and  is  equal  (8.  Ax.)  to  it. 

PROP.  IX.     PROB. 


To  bisect  a  given  rectilineal  angle,  that  is,  to  divide  it  into  two  equal  angles. 

Let  BAG  be  the  given  rectilineal  angle,  it  is  required  to  bisect  it. 

Take  any  point  D  in  AB,  and  from  AG  cut 
(3.  1.)  off  AE  equal  to  AD  ;  join  DE,  and  upon 
it  describe  (1.1.)  an  equilateral  triangle  DEF  ; 
then  join  AF  ;  the  straight  line  AF  bisects 
the  angle  BAG. 

Because  AD  is  equal  to  AE,  and  AF  is  com- 
mon to  the  two  triangles  DAF,  E  AF  ;  the  two 
sides  DA,  AF,  are  equal  to  the  two  sides  EA, 
AF,  each  to  each ;  but  the  base  DF  is  also 
equal  to  the  base  EF  ;  therefore  the  angle 
DAF  is  equal  (8. 1 .)  to  the  angle  E  AF  :  where- 
fore the  given  rectilineal  angle  BAG  is  bisect- 
ed by  the  straight  line  AF. 

SGHOLIUM. 

By  the  same  construction,  each  of  the  halves  BAF,  GAF,  may  be  divi- 
ded into  two  equal  parts  ;  and  thus,  by  successive  subdivisions,  a  given  an- 
gle may  be  divided  into  four  equal  parts,  into  eight,  into  sixteen,  and  so  on. 

PROP.  X.     PROB. 

•iw*      To  bisect  a  given  finite  straight  line,  that  is,  to  divide  it  into  two  equal  parts. 

Let  AB  be  the  given  straight  line  ;  it  is  required  to  divide  it  into  two  equal 
parts. 

Describe  (1.  1.)  upon  it  an  equilateral  triangle  ABG,  and  bisect  (9.  1.) 
the  angle  ACB  by  the  straight  line  GD.     AB  is 
cut  into  two  equal  parts  in  the  point  D. 

Because  AG  is  equal  to  GB,  and  GD  common 
to  the  two  triangles  AGD,  BGD  :  the  two  sides 
AC,  CD,  are  equal  to  the  two  BC,  CD,  each  to 
each  ;  but  the  angle  AGD  is  also  equal  to  the  an- 
gle BGD ;  therefore  the  base  AD  is  equal  to  the 
base  (4.  1.)  DB,  and  the  straight  line  AB  is  divi- 
ded into  two  equal  parts  in  the  point  D. 


18 


ELEMENTS 


PROP.  XI.     PROB. 

To  draw  a  straight  line  at  right  angles  to  a  given  straight  line,  from  a  given 
point  in  that  line. 

Let  AB  be  a  given  straight  line,  and  C  a  point  given  in  it ;  it  is  requi- 
red to  draw  a  straight  line  from  the  point  C  at  right  angles  to  AB. 

Take  any  point  D  in  AC,  and  (3.  I.)  make  CE  equal  to  CD,  and  upon 
DE   describe  (1.  1.)  the  equilateral  -p 

triangle  DFE,  and  join  FC;  the 
straight  line  FC,  drawn  from  the  giv- 
en point  C,  is  at  right  angles  to  the 
given  straight  line  AB. 

Because  DC  is  equal  to  CE,  and 
FC  common  to  the  two  triangles 
DCF,  EOF,  the  two  sides  DC,  CF 

are  equal  to  the  two  EC,  CF,  each       ADC  E        B 

to  each;  but  the  base  DF  is  also  equal  to  the  base  EF;  therefore  the 
angle  DCF  is  equal  (8.  1.)  to  the  angle  ECF  ;  and  they  are  adjacent  an- 
gles. But,  when  the  adjacent  angles  which  one  straight  line  makes  with 
another  straight  line  are  equal  to  one  another,  each  of  them  is  called  a 
right  (7.  def.)  angle  ;  therefore  each  of  the  angles  DCF,  ECF,  is  a  right 
angle.  Wherefore,  from  the  given  point  C,  in  the  given  straight  line  AB, 
FC  has  been  drawn  at  right  angles  to  AB. 

PROP.  XIL     PROB. 

To  draw  a  straight  line  perpendicular  to  a  given  straight  line,  of  an  unlimited 
length,  from  a  given  point  without  it.     . 

^  I^et  AB  be  a  given  straight  line,  which  may  be  produced  to  any  length 
both  ways,  and  let  C  be  a  point  without  it.  It  is  required  to  draw  a  straight 
line  perpendicular  to  AB  from  the 
point  C. 

Take  any  point  D  upon  the  other 
side  of  ABjand  from  the  centre  C,  at 
the  distance  CD,  describe  (3.  Post.) 
the  circle  EOF  meeting  AB  in  F,  G : 
and  bisect  (10.  1.)  FG  in  H,  and  join 
CF,  CH,  CG  ;  the  straight  line  CH, 
drawn  from  the  given  point  C,  is  per- 
pendicular to  the  given  straight  line  AB. 

Because  FH  is  equal  to  HG,  and  HC  common  to  the  two  triangles  FHC, 
GHC,  the  two  sides  FH,  HC  are  equal  to  the  two  GH,  HC,  each  to  each , 
but  the  base  CF  is  also  equal  (11.  Def.  1.)  to  the  base  CG ;  therefore  the 
angle  CHF  is  equal  (8.  1.)  to  the  angle  CHG  ;  and  they  are  adjacent  an- 
gles ;  now  when  a  straight  line  standing  on  a  straight  line  makes  the  ad- 
jacent angles  equal  to  one  another,  each  of  them  is  a  right  angle,  and 
the  straight  line  which  stands  upon  the  other  is  called  a  perpendicular  to 
it ;  therefore  from  the  given  point  C  a  perpendicular  CH  has  been  drawn 
to  the  given  straight  line  AB. 


OF  GEOMETRY.    BOOK  I. 


19 


PROP.  XIII.    THEOR. 

The  angles  which  one  straight  line  makes  with  another  upon  one  side  of  it,  are 
either  two  right  angles,  or  are  together  equal  to  two  right  angles. 

Let  the  straight  line  AB  make  with  CD,  upon  one  side  of  it  the  angles 
CBA,  ABD  ;  these  are  either  two  right  angles,  or  are  together  equal  to  two 
right  angles. 

For,  if  the  angle  CBA  be  equal  to  ABD,  each  of  them  is  a  right  angle 
(Def.  7.) ;  but,  if  not,  from  the  point  B  draw  BE  at  right  angles  (11.  1.) 


B  0  fi  B  5 

to  CD  ;  therefore  the  angles  CBE,  EBD  are  two  right  angles.  Now,  the 
angle  CBE  is  equal  to  the  two  angles  CBA,  ABE  together;  add  the  an- 
gle EBD  to  each  of  these  equals,  and  the  two  angles  CBE,  EBD  will  be 
equal  (2.  Ax.)  to  the  three  CBA,  ABE,  EBD.  Again,  the  angle  DBA  is 
equal  to  the  two  angles  DBE,  EBA  ;  add  to  each  of  these  equals  the  angle 
ABC  ;  then  will  the  two  angles  DBA,  ABC  be  equal  to  the  three  angles 
DBE,  EBA,  ABC ;  but  the  angles  CBE,  EBD  have  been  demonstrated 
to  be  equal  to  the  same  three  angles  ;  and  things  that  are  equal  to  the  same 
are  equal  (1.  Ax.)  to  one  another;  therefore  the  angles  CBE,  EBD  are 
equal  to  the  angles  DBA,  ABC  ;  but  CBE,  EBD,  are  two  right  angles  ; 
therefore  DBA,  ABC  ;  are  together  equal  to  two  right  angles. 

CoR.  The  sum  of  all  the  angles,  formed  on  the  same  side  of  a  straight 
line  DC,  is  equal  to  two  right  angles  ;  because  their  sum  is  equal  to  that 
of  the  two  adjacent  angles  DBA,  ABC. 

*      PROP.  XIV.    THEOR. 

If,  at  a  point  in  a  straight  line,  two  other  straight  lines,  upon  the  opposite 
sides  of  it,  make  the  adjacent  angles  together  equal  to  two  right  angles, 
these  two  straight  lines  are  in  one  and  the  same  straight  line. 

At  the  point  B  in  the  straight  line  AB, 
let  the  two  straight  lines  BC,  BD  upon 
the  opposite  sides  of  AB,  make  the  adja- 
cent angles  ABC,  ABD  equal  togethe* 
to  two  right  angles.  BD  is  in  the  same 
straight  line  with  CB. 

For  if  BD  be  not  in  the  same  straight 
line  with  CB,  let  BE  be  in  the  same 
straight  line  with  it ;  therefore,  because 
the  straight  line  AB  makes  angles  with 
the  straight  line  CBE,  upon  one  side  of 


20  ELEMENTS 

it,  t^ae  angles  ABC,  ABE  are  together  equal  (13.  1.)  to  two  right  angles ; 
but  the  angles  ABC,  ABD  are  likewise  together  equal  to  two  right  angles  : 
therefore  the  angles  CBA,  ABE  are  equal  to  the  angles  CBA,  ABD  : 
Take  away  the  common  angle  ABC,  and  the  remaining  angle  ABE  is  equal 
(3.  Ax.)  to  the  remaining  angle  ABD,  the  less  to  the  greater,  which  is  im- 
possible ;  therefore  BE  is  not  in  the  same  straight  line  with  BC.  And  ia 
like  manner,  it  may  be  demonstrated,  that  no  other  can  be  in  the  same 
straight  line  with  it  but  BD,  which  therefore  is  in  the  same  straight  line 
with  CB. 

PROP.  XV.    THEOR. 

If  two  straight  lines  cut  one  another^  the  vertical^  or  opposite  angles  are  equal. 

Let  the  two  straight  lines  AB,  CD,  cut  one  another  in  the  point  E  :  the 
angle  AEC  shall  be  equal  to  th&  angle  DEB,  and  CEB  to  AED. 

For  the  angles  CEA,  AED,  which  the  straight  line  AE  makes  with  the 
straigiht  line  CD,  are  together  equal  (13. 1.)  to  two  right  angles  :  and  the 
angles    AED,  DEB,   which    the 
straight  line  DE  makes  with  the 
straight  line  AB,  are  also  together 
equal  (13.  1.)  to  two  right  angles  ; 
therefore   the   two   angles    CEA, 
AED  are  equal  to  the  two  AED, 
DEB.     Take  away  the  common 
angle    AED,   and   the   remaining 
angle  CEA  is  equal  (3.  Ax.)  to  the 
remaining   angle    DEB.      In   the 
same  manner  it  may  be   demonstrated  that  the  angles  CEB,  AED  are 
equal. 

Cor.  1.  From  this  it  is  manifest,  that  if  two  straight  lines  cut  one  an- 
other, the  angles  which  they  make  at  the  point  of  their  intersection,  are 
together  equal  to  four  right  angles. 

CoR.  2.  And  hence,  all  the  angles  made  by  any  number  of  straight  lines 
meeting  in  one  point,  are  together  equal  to  four  right  angles. 

PROP.  XVL    THEOR. 

If  one  side  of  a  triangle  he  produced,  the  exterior  angle  is  greater  than 
either  of  the  interior ,  and  opposite  angles. 

Let  ABC  be  a  triangle,  and  let  its  side  BC  be  produced  to  D,  the  ex- 
terior angle  ACD  is  greater  than  either  of  the  interior  opposite  angles 
GBA,  BAC. 

Bisect  (10.  1.)  AC  in  E,  join  BE  and  produce  it  to  F,  and  make  EF 
equal  to  BE  ;  join  also  FC,  and  produce  AC  to  G. 

Because  AE  is  equal  to  EC,  and  BE  to  EF  ;  AE,  EB  are  equal  to 
CE,  EF,  each  to  each;  and  the  angle  AEB  is  equal  (15.  1.)  to  the 
angle  CEF,  because  they  are  vertical  angles ;  therefore  the  base  AB 


OF  GEOMETRY,    BOOK  I. 


21 


is  equal  (4.  1.)  to  the  base  CF,  and 
the  triangle  AEB  to  the  triangle 
CEF,  and  the  remaining  angles  to 
the  remaining  angles  each  to  each, 
to  which  the  equal  sides  are  oppo- 
site ;  wherefore  the  angle  BAE  is 
equal  to  the  angle  ECF ;  but  the 
angle  ECD  is  greater  than  the  an- 
gle ECF  ;  therefore  the  angle  ECD, 
that  is  ACD,  is  greater  than  BAE : 
In  the  same  manner,  if  the  side  BC 
be  bisected,  it  may  be  demonstrated 
that  the  angle  BCG,  that  is  (15.  1.), 
the  angle  ACD,  is  greater  than  the 
angle  ABC. 


PROP.  XVII.    THEOR. 
Any  tvffo  angles  of  a  triangle  are  together  less  than  two  right  angles. 

Let  ABC  be  any  triangle ;  any 
two  of  its  angles  together  are  less 
than  two  right  angles. 

Produce  BC  to  D ;  *and  because 
ACD  is  the  exterior  angle  of  the  tri- 
angle ABC,  ACD  is  greater  (16.  1.) 
than  the  interior  and  opposite  angle 
ABC  ;  to  each  of  these  add  the  angle 
ACB ;  therefore  the  angles  ACD, 
ACB  are  greater  than  the  angles 
ABC,  ACB  ;  but  ACD,  ACB  are  to- 
gether equal  (13.  1.)  to  two  right  an- 
gles :  therefore  the  angles  ABC,  BCA  are  less  than  two  right  angles.  In 
like  manner,,  it  may  be  demonstrated,  that  BAC,  ACB  as  also  CAB,  ABC, 
are  less  than  two  right  angles. 

PROP.  XVIII.    THEOR. 

The  greater  side  of  every  triangle  has  the  greater  angle  opposite  to  it. 

Let  ABC  be  a  triangle  of  which  the 
side  AC  is  greater  than  the  side  AB  ;  the 
angle  ABC  is  also  greater  than  the  angle 
BCA. 

From  AC,  which  is  greater  than  AB, 
cut  off  (3.  1.)  AD  equal  to  AB,  and  join 
BD :  and  because  ADB  is  the  exterior 
angle  of  the  triangle  B DC,  it  is  greater 
(16.  1.)  than  the  interior  and  opposite 


23  ELEMENTS 

angle  DCB ;  but  ADB  is  equal  (5.  1.)  to  ABD,  because  the  side  AB  is 
equal  to  the  side  AD  ;  therefore  the  angle"  ABD  is  likewise  greater  than 
the  angle  ACB ;  wherefore  much  more  is  the  angle  ABC  greater  than 
ACB 

PROP.  XIX.     THEOR. 

The  greater  angle  of  every  triangle  is  subtended  hy  the  greater  side,  or  has 
the  greater  side  opposite  to  it. 

Let  ABG  be  a  triangle,  of  which  the  angle  ABC  is  greater  than  the 
angle  BCA ;  the  side  AC  is  likewise  greater  than  the  side  AB. 

For,  if  it  be  not  greater,  AC  must  either 
be  equal  to  AB,  or  less  than  it ;  it  is  not 
equal,  because  then  the  angle  ABC  would 
be  equal  (5.  1.)  to  the  angle  ACB  ;  but  it  is 
not ;  therefore  AC  is  not  equal  to  AB  ;  nei- 
ther is  it  less  ;  because  then  the  angle  ABC 
would  be  less  (18.  l.)than  the  angle  ACB  ; 
but  it  is  not ;  therefore  the  side  AC  is  not  -r* 
less  than  AB  ;  and  it  has  been  shewn  that 
it  is  not  equal  to  AB  ;  therefore  AC  is  greater  than  AB. 

PROP.  XX.     THEOR. 
Any  two  sides  of  a  triangle  are  together  greater  than  the  third  side. 

Let  ABC  be  a  triangle  ;  any  two  sides  of  it  together  arc  greater  than 
the  third  side,  viz.  the  sides  BA,  AC  greater  than  the  side  BC  ;  and  AB, 
BC  greater  than  AC  ;  and  BC,  CA  greater  than  AB. 

Produce  BA  to  the  point  D,  and  make 
(3.  1.)  AD  equal  to  AC  ;  and  join  DC. 

Because  DA  is  equal  to  AC,  the  an-  . 

gle  ADC  is  likewise  equal    (5.  1.)  to  ^ 

ACD  ;  but  the  angle  BCD  is  greater 
than  the  angle  ACD  ;  therefore  the  an- 
gle BCD  is  greater  than  the  angle 
ADC ;  and  because  the  angle  BCD  of 
the  triangle  DCB  is  greater  than  its  an-      B  O 

gle  BDC,  and  that  the  greater  (19.  1.)  side  is  opposite  to  the  greater  an- 
gle ;  therefore  the  side  DB  is  greater  than  the  side  BC  ;  but  DB  is  equal 
to  BA  and  AC  together ;  therefore  BA  and  AC  together  are  greater  than 
BC.  In  the  same  manner  it  may  be  demonstrated,  that  the  sides  AB, 
BC  are  greater  than  CA,  and  BC,  CA  greater  than  AB. 

SCHOLIUM. 

This  may  be  demonstrated  without  producing  any  of  the  sides  :  thus, 
the  line  BC,  for  example,  is  the  shortest  distance  from  B  to  C ;  therefore 
BC  is  less  than  BA+AG  or  BA+AC>BC. 


OF  GEOMETRY.    BOOK  I.  y 


^ 


PROP.  XXI.     THEOR. 

If  from  the  ends  of  one  side  of  a  triangle,  there  be  drawn  two  straight 
lines  to  a  point  within  the  triangle,  these  two  lines  shall  be  less  than  the 
other  two  sides  of  the  triarigle,  but  shall  contain  a  greater  angle. 

Let  the  two  straight  lines  BD,  CD  be  drawn  from  B,  C,  the  ends  of 
the  side  BC  of  the  triangle  ABC,  to  the  point  D  within  it;  BD  and  DC 
are  less  than  the  other  two  sides  BA,  AC  of  the  triangle,  but  contain  an 
angle  BDC  greater  than  the  angle  BAC. 

Produce  BD  to  E  ;  and  because  two  sides  of  a  triangle  (20.  1.)  are 
greater  than  the  third  side,  the  two  sides  B  A, 
AE  of  the  triangle  ABE  are  greater  than  BE. 
To  each  of  these  add  EC ;  therefore  the 
sides  BA,  AC  are  greater  than  BE,  EC  ; 
Again,  because  the  two  sides  CE,  ED,  of 
the  triangle  CED  are  greater  than  CD,  if 
DB  be  added  to  each,  the  sides  CE,  EB, 
will  be  greater  than  CD,  DB ;  but  it  has 
been  shewn  that  BA,  AC  are  greater  than 
BE,  EC  ;  much  more  then  are  BA,  AC  great- 
er than  BD,  DC. 

Again,  because  the  exterior  angle  of  a 
triangle  (16.  1.)  is  greater  than  the  interior  and  opposite  angle,  the  exte- 
rior angle  BDC  of  the  triangle  CDE  is  greater  than  CED  ;  for  the  same 
reason,  the  exterior  angle  CEB  of  the  triangle  ABE  is  greater  than  BAC  ; 
and  it  has  been  demonstrated  that  the  angle  BDC  is  greater  than  the 
angle  CEB ;  much  more  then  is  the  angle  BDC  greater  than  the  angle 
BAC, 

PROP.  XXII.     PROB. 

To  construct  a  triangle  of  which  the  sides  shall  be  equal  to  three  given 
straight  lines ;  but  any  two  whatever  of  these  lines  must  be  greater  than 
the  third  (20.  1.), 

Let  A,  B,  C  be  the  three  given 
straight  lines,  of  which  any  two 
whatever  are  greater  than  the 
third,  viz.  A  and  B  greater  than 
C  ;  A  and  C  greater  than  B  ;  and 
B  and  C  than  A.  It  is  required 
to  make  a  triangle  of  which  the 
sides  shall  be  equal  to  A,  B,  C, 
each  to  each. 

Take  a  straight  line  DE,  ter- 
minated at  the  point  D,  but  un- 
limited towards  E,  and  make 
(3.  1.)  DF  equal  to  A,  FG  to  B, 
and  GH  equal  to  C ;  and  from 


24  ELEMENTS 

the  centre  F,  at  the  distance  FD,  describe  (3.  Post.)  the  circle  DKL ; 
and  from  the  centre  G,  at  the  distance  GH,  describe  (3.  Post.)  another 
circle  HLK ;  and  join  KF,  KG ;  the  triangle  KFG  has  its  sides  equal  to 
the  three  straight  lines,  A,  B,  0. 

Because  the  point  F  is  the  centre  of  the  circle  DKL,  FD  is  equal  (11. 
Def.)  to  FK ;  but  FD  is  equal  to  the  straight  line  A  ;  therefore  FK  is 
equal  to  A:  Again,  because  G  is  the  centre  of  the  circle  LKH,  GH  is 
equal  (11.  Def)  to  GK ;  but  GH  is  equal  to  C;  therefore,  also,  GK  is 
equal  to  C  ;  and  FG  is  equal  to  B  ;  therefore  the  three  straight  lines  KF, 
FG,  GK,  are  equal  to  the  three  A,  B,  C :  And  therefore  the  triangle 
KFG  has  its  three  sides  KF,  FG,  GK  equal  to  the  three  given  straight 
lines,  A,  B  C. 

SCHOLIUM. 

If  one  of  the  sides  were  greater  than  the  sum  of  the  other  two,  the  arcs 
would  not  intersect  each  other :  but  the  solution  will  always  be  possible, 
when  the  sum  of  two  sides,  any  how  taken  (20.  1.)  is  greater  than  the 
third. 

PROP.  XXIII.     PROB. 

At  a  given  point  in  a  given  straight  line,  to  make  a  rectilineal  angle  equal 
to  a  given  rectilineal  angle. 

Let  AB  be  the  given  straight  line,  and  A  the  given  point  in  it,  and  DOE 
the  given  rectilineal  angle ;  it  is  required  to  make  an  angle  at  the  given 
point  A  in  the  given  straight  line 
AB,  that  shall  be  equal  to  the 
given  rectilineal  angle  DOE. 

Take  in  CD,  CE  any  points  D, 
E,  and  join  DE  ;  and  make  (22. 
1.)  the  triangle  AFG,  the  sides 
of  which  shall  be  equal  to  the 
three  straight  lines,  CD,  DE,  CE, 
so  that  CD  be  equal  to  AF,  CE  to 
AG,  and  DE  to  FG  ;  and  because 
DC,  CE  are  equal  to  FA,  AG, 
each  to  each,  and  the  base  DE  to 
the  base  FG ;  the  angle  DCE  is 
equal  (8.  1.)  to  the  angle  FAG. 
Therefore,  at  the  given  point  A  in  the  given  straight  line  AB,  the  angle 
FAG  is  made  equal  to  the  given  rectilineal  angle  DCE. 

PROP.  XXIV.    THEOR. 

If  two  triangles  have  two  sides  of  the  one  equal  to  two  sides  of  the  other,  each 
to  each,  but  the  angle  contained  hy  the  two  sides  of  the  one  greater  than 
the  angle  contained  hy  the  two  sides  of  the  other  ;  the  base  of  that  which 
has  the  greater  angle  shall  be  greater  than  the  base  of  the  other. 

Let  ABC,  DEF  be  two  triangles  which  have  the  two  sides  AB,  AC 
equal  to  the  two  DE,  DF  each  to  each,  viz.  AB  equal  to  DE,  and  AC  to 


OF  GEOMETRY.    BOOK  I. 


25 


DF ;  but  the  angle  BAG  greater  than  the  angle  EDF ;  the  base  BG  is 
also  greater  than  the  base  EF. 

Of  the  two  sides  DE,  DF,  let  DE  be  the  side  which  is  not  greater  than 
the  other,  and  at  the  point  D,  in  the  straight  line  DE,  make  (23,  1.)  the 
angle  EDG  equal  to  the  angle  BAG :  and  make  DG  equal  (3.  1.)  to  AG 
or  DF,  andjoin  EG,  GF. 

Because  AB  is  equal  to  DE,  and  AG  to  DG,  the  two  sides  BA,  AG  are 
equal  to  the  two  ED,  DG,  each  to  each,  and  the  angle  BAG  is  equal  to 
the  angle  EDG,  therefore 

the  base  BG  is  equal  (4. 1.)  A  O 

to  the  base  EG;  and  be- 
cause DG  is  equal  to  DF, 
the  angle  DFG  is  equal 
(5.  1.)  to  the  angle  DGF; 
but  the  angle  DGF  is 
greater  than  the  angle 
EGF ;  therefore  the  angle 
DFG  is  greater  than  EGF; 
and  much  more  is  the  angle 
EFG  greater  than  the 
angle  EGF ;  and  because 
the  angle  EFG  of  the  triangle  EFG  is  greater  than  its  angle  EGF,  and 
because  the  greater (19.  l.)side  is  opposite  to  the  greater  angle,  the  side 
EG  is  greater  than  the  side  EF ;  but  EG  is  equal  to  BG  ;  and  therefore 
also  BG  is  greater  than  EF. 


PROP.  XXV.    THEOR. 

If  two  triangles  have  two  sides  of  the  one  equal  to  two  sides  of  the  other,  each 
to  each,  but  the  base  of  the  otie  greater  than  the  base  of  the  other ;  the  angle 
contained  by  the  sides  of  that  which  has  the  greater  base,  shall  be  greater 
than  the  angle  contained  by  the  sides  of  the  other. 

Let  ABG,  DEF  be  two  triangles  which  have  the  two  sides,  AB,  AG, 
equal  to  the  two  sides  DE,  DF,  each  to  each,  viz.  AB  equal  to  DE,  and 
AC  to  DF  :  but  let  the  base  CB  be  greater  than  the  base  EF,  the  angle 
BAG  is  likewise  greater  than  the  angle  EDF. 

For,  if  it  be  not  greater,  it  must  either  be  equal  to  it,  or  less ;  but  the 
angle  Bx\C  is  not  equal  to  the  angle 
EDF,  because  then  the  base  BG 
would  be  equal  (4. 1 .)  to  EF  ;  but  it  is 
not ;  therefore  the  angle  BAG  is  not 
equal  to  the  angle  EDF  ;  neither  is 
it  less  ;  because  then  the  base  BG 
would  be  less  (24.  1.)  than  the  bas'^ 
EF  ;  but  it  is  not ;  therefore  the  an- 
gle BAG  is  not  less  than  the  angle 
EDF :  and  it  was  shewn  that  it  is 
nor  equal  to  it :  therefore  the  angle 
BAG  is  greater  than  the  angle  EDF. 


26 


ELEMENTS 


PROP.  XXVI.     THEOR. 

If  two  triangles  have  two  angles  of  the  one  equal  to  two  angles  of  the  other ^ 
each  to  each ;  and  one  side  equal  to  one  side,  viz.  either  the  sides  adjacent 
to  the  equal  angles ^  or  the  sides  opposite  to  the  equal  angles  in  each ;  then 
shall  the  other  side  be  equals  each  to  each ;  and  also  the  third  angle  of  the 
one  to  the  third  angle  of  the  other. 

Let  ABC,  DEF  be  two  trian- 
gles which  have  the  angles 
ABC,  BCA  equal  to  the  angles 
DEF,  EFD,  viz.  ABC  to  DEF, 
and  BCA  to  EFD,  also  one  side 
equal  to  one  side  ;  and  first,  let 
those  sides  be  equal  which  are 
adjacent  to  the  angles  that  are 
equal  in  the  two  triangles,  viz. 
BC    to    EF ;  the   other    sides 

shall  be  equal,  each  to  each,  viz.  rt     -ni  -rt 

AB  to  DE,  and  AC  to  DF  ;  and      B  t.     JE  h 

the  third  angle  BAC  to  the  third  angle  EDF. 

For,  if  AB  be  not  equal  to  DE,  one  of  them  must  be  the  greater.  Let 
AB  be  the  greater  of  the  two,  and  make  BG  equal  to  DE,  and  join  GC  ; 
therefore,  because  BG  is  equal  to  DE,  and  BC  to  EF,  the  two  sides  GB, 
BC  are  equal  to  the  two,  DE,  EF,  each  to  each ;  and  the  angle  GBC  is 
equal  to  the  angle  DEF ;  therefore  the  base  GC  is  equal  (4.  1.)  to  the 
base  DF,  and  the  triangle  GBC  to  the  triangle  DEF,  and  the  other  angles 
to  the  other  angles,  each  to  each,  to  which  the  equal  sides  are  opposite  ; 
therefore  the  angle  GCB  is  equal  to  the  angle  DFE,  but  DFE  is,  by  the 
hypothesis,  equal  to  the  angle  BCA ;  wherefore  also  the  angle  BCG  is 
equal  to  the  angle  BCA,  the  less  to  the  greater,  which  is  impossible  ; 
therefore  AB  is  not  unequal  to  DE,  that  is,  it  is  equal  to  it ;  and  BC  is 
equal  to  EF  ;  therefore  the  two  AB,  BC  are  equal  to  the  two  DE,  EF, 
each  to  each  ;  and  the  angle  ABC  is  equal  to  the  angle  DEF ;  therefore 
the  base  AC  is  equal  (4.  1.)  to  the  base  DF,  and  the  angle  BAC  to  the 
angle  EDF. 

Next,  let  the  sides  which  are 
opposite  to  equal  angles  in  each 
triangle  be  equal  to  one  another, 
viz.  AB  to  DE  ;  likewise  in  this 
case,  the  other  sides  shall  be 
equal,  AC  to  DF,  and  BC  to  EF  ; 
and  also  the  third  angle  BAC  to 
the  third  EDF. 

For,  if  BC  be  not  equal  to  EF, 
let  BC  be  the  greater  of  them, 
and  make  BH  equal  to  EF,  and 
join  AH  ;  and  because  BH  is 
equal  to  EF,  and  AB  to  DE  ;  the  two  AB,  BH  aw  equal  to  the  two 
DE,  EF,  each  to  each;  and  they  contain  equal  angles;  therefore  (4.  1.) 


OF  GEOMETRY.    BOOK  I. 


27 


the  base  AH  is  equal  to  the  base  DF,  and  the  triangle  ABH  to  the  trian- 
gle DEF,  and  the  other  angles  are  equal,  each  to  each,  to  which  the  equal 
sides  are  opposite  ;  therefore  the  angle  BHA  is  equal  to  the  angle  EFD; 
but  EFD  is  equal  to  the  angle  BCA  ;  therefore  also  the  angle  BHA  is  equal 
to  the  angle  BCA,  that  is,  the  exterior  angle  BHA  of  the  triangle  AHC  is 
equal  to  its  interior  and  opposite  angle  BCA,  which  is  impossible  (16.  1.) ; 
wherefore  BC  is  not  unequal  to  EF,  that  is,  it  is  equal  to  it ;  and  AB  is 
equal  to  DE ;  therefore  the  two,  AB,  BC  are  equal  to  the  two  DE,  EF,  each 
to  each  ;  and  they  contain  equal  angles  ;  wherefore  the  base  AC  is  equal 
to  the  base  DF,  and  the  third  angle  BAC  to  the  third  angle  EDF. 

PROP.  XXVn.     THEOR. 


If  a  straight  line  falling  upon  two  other  straight  lines  makes  the  alternate 
angles  equal  to  one  another j  these  two  straight  lines  are  parallel. 

Let  the  straight  line  EF,  which  falls  upon  the  two  straight  lines  AB, 
CD  make  the  alternate  angles  AEF,  EFD  equal  to  one  another ;  AB  is 
parallel  to  CD. 

For,  if  it  be  not  parallel,  AB  and  CD  being  produced  shall  meet  either 
towards  B,  D,  or  towards  A,  C  ;  let  them  be  produced  and  meet  towards 
B,  D  in  the  point  G ;  therefore  GEF  is  a  triangle,  and  its  exterior  angle 
AEF  is  greater  (16-  1.)  than  the  interior  and  opposite  angle  EFG  ;  but  it 
is  also  equal  to  it,  which  is  im- 
possible :  therefore,  AB  and  CD 
being  produced,  do  not  meet  to- 
wards B,  D.  In  like  manner  it 
may  be  demonstrated  that  they 
do  not  meet  towards  A,  C  ;  but 
those  straight  lines  which  meet 
neither  way,  though  produced 
ever  so  far,  are  parallel  (30.  Def.) 
to  one  another.     AB  therefore  is  parallel  to  CD. 

PROP.  XXVni.     THEOR. 

If  a  straight  line  falling  upon  two  other  straight  lines  makes  the  exterior  an- 
gle equal  to  the  interior  and  opposite  upon  the  same  side  of  the  line  ;  or 
makes  the  interior  angles  upon  the  same  side  together  equal  to  two  right 
angles ;  the  two  straight  lines  are  parallel  to  one  another. 

Let  the  straight  line  EF,  which 
falls  upon  the  two  straight  lines  AB, 
CD,  make  the  exterior  angle  EGB 
equal  to  GHD,  the  interior  and  oppo- 
site angle  upon  the  same  side  ;  or  let  it 
make  the  interior  angles  on  the  same 
side  BGH,  GHD  together  equal  to  two 
right  angles  ;  AB  is  parallel  to  CD. 

Because  the  angle  EGB  is  equal  to 
the  angle  GHD,  and  also  (15.  1.)  to  the 


c 


H^ 


28  ELEMENTS 

angle  AGH,  the  angle  AGH  is  equal  to  the  angle  GHD  ;  and  they  are  th* 
alternate  angles  ;  therefore  AB  is  parallel  (27.  1.)  to  CD.  Again,  becaussj 
the  angles  BGH,  GHD  are  equal  (hyp. )to  two  right  angles,  and  x\GH,  BGH, 
are  also  equal  (13.  1.)  to  two  right  angles,  the  angles  AGH,  BGH  are  equal 
to  the  angles  BGH,  GHD  :  Take  away  the  common  angle  BGH  ;  therefore 
the  remaining  angle  AGH  is  equal  to  the  remaining  angle  GHD  ;  and  they 
are  alternate  angles  ;  therefore  AB  is  parallel  to  CD. 

CoR.  Hence,  when  two  straight  lines  are  perpendicular  to  a  third  line, 
they  will  be  parallel  to  each  other. 

PROP.  XXIX.    THEOR. 

If  a  straight  line  fall  upon  two  parallel  straight  lines,  it  makes  the  alternate 
angles  equal  to  one  another ;  and  the  exterior  angle  equal  to  the  interior 
and  opposite  upon  the  same  side ;  and  likewise  the  two  interim"  angles  upon 
the  same  side  together  equal  to  two  right  angles. 

Let  the  straight  line  EF  fall  upon  the  parallel  straight  lines  AB,  CD  ; 
the  alternate  angles  AGH,  GHD  are  equal  to  one  another ;  and  the  exte- 
rior angle  EGB  is  equal  to  the  interior  and  opposite,  upon  the  same  side, 
GHD ;  and  the  two  interior  angles  BGH,  GHD  upon  the  same  side  are 
together  equal  to  two  right  angles. 

For  if  AGH  be  not  equal  to  GHD,  let  KG  be  drawn  making  the  angle 
KGH  equal  to  GHD,  and  produce  KG  to  L ;  then  KL  will  be  parallel  to 
CD  (27. 1.) ;  but  AB  is  also  paral- 
lel to  CD  ;  therefore  two  straight 
lines  are  drawn  through  the  same 
point  G,  parallel  to  CD,  and  yet 
not  coinciding  with  one  another, 
which  is  impossible  (11.  Ax.)  The 
angles  AGH,  GHD  therefore  are 
not  unequal,  that  is,  they  are  equal 
to  one  another.  Now,  the  angle 
EGB  is  equal  to  AGH  (15.  1.) ; 
and  AGH  is  proved  to  be  equal 
to  GHD  ;  therefore  EGB  is  like- 
wise equal  to  GHD  ;  add  to  each  of  these  the  angle  BGH  ;  therefore  the 
angles  EGB,  BGH  are  equal  to  the  angles  BGH,  GHD  ;  but  EGB,  BGH 
are  equal  (13.  1.)  to  two  right  angles;  therefore  also  BGH,  GHD  are 
equal  to  two  right  angles. 

Cor.  1.  If  two  lines  KL  and  CD  make,  with  EF,  the  two  angles  KGH, 
GHC  together  less  than  two  right  angles,  KG  and  CH  will  meet  on  the  side 
of  EF  on  which  the  two  angles  are  that  are  less  than  two  right  angles. 

For,  if  not,  KL  and  CD  are  either  parallel,  or  they  meet  on  the  other 
side  of  EF  ;  but  they  are  not  parallel ;  for  the  angles  KGH,  GHC  would 
then  be  equal  to  two  right  angles.  Neither  do  they  meet  on  the  other 
side  of  EF ;  for  the  angles  LGH,  GHD  would  then  be  two  angles  of  a 
triangle,  and  less  than  two  right  angles ;  but  this  is  impossible ;  for  the 
four  angles  KGH,  HGL,  CHG,  GHD  are  together  equal  to  four  right 
angles  (13.  1.)  of  which  the  two,  KGH,  CHG,  are  by  supposition  less  than 


OF  GEOMETRY.    BOOK  I.  29 

» 

two  right  angles ;  therefore  the  other  two,  HGL,  GHD  are  greater  than 
two  right  angles.  Therefore,  since  KL  and  CD  are  not  parallel,  and  since 
they  do  not  meet  towards  L  and  D,  they  must  meet  if  produced  towards 
K  and  0. 

Cor.  2.  If  BGH  is  a  right  angle,  GHD  will  be  a  right  angle  also ; 
therefore  every  line  perpendicular  to  one  of  two  parallels,  is  perpendicular 
to  the  other. 

CoR.  3.  Since  AGE^BGH,  and  DHF=;CHG ;  hence  the  four  acute 
angles  BGH,  AGE,  GHC,  DHF,  are  equal  to  each  other.  The  same  is 
the  case  with  the  four  obtuse  angles  EGB,  AGH,  GHD,  CHF.  It  may 
be  also  observed,  that,  in  adding  one  of  the  acute  angles  to  one  of  the  ob- 
tuse, the  sum  will  always  be  equal  to  two  right  angles. 

SCHOLIUM.  • 

The  angles  just  spoken  of,  when  compared  with  each  other,  assume 
different  names.  BGH,  GHD,  we  have  already  named  interior  angles  on 
the  same  side ;  AGH,  GHC,  have  the  same  name  ;  AGH,  GHD,  are  called 
alternate  interior  angles,  or  simply  alternate ;  so  also,  are  BGH,  GHC  : 
and  lastly,  EGB,  GHD,  or  EGA,  GHC,  are  called,  respectively,  the  op- 
posite exterior  and  interior  angles  ;  and  EGB,  CHF,  or  AGE,  DHF,  the 
alternate  exterior  angles. 

PROP.  XXX.     THEOR. 

Straight  lines  which  are  parallel  to  the  same  straight  line  are  parallel  to  one 

another. 

Let  AB,  CD,  be  each  of  them  parallel  to  EF  ;  AB  is  also  parallel  to 
CD. 

Let  the  straight  line  GHK  cut  AB,  EF,  CD  ;  and  because  GHK  cuts 
the  parallel  straight  lines  AB,  EF,  the 
angle  AGH  is  equal  (29.  1.)  to  the  an- 
gle GHF.  Again,  because  the  straight 
line  GK  cuts  the  parallel  straight  lines 
EF,  CD,  the  angle  GHF  is  equal  (29. 
1.)  to  the  angle  GKD :  and  it  was 
shewn  that  the  angle  AGK  is  equal  to 
the  angle  GHF ;  therefore  also  AGK 
is  equal  to  GKD  ;  and  they  are  alter- 
nate angles ;  therefore  AB  is  parallel 
(27.  1.)  to  CD. 

PROP.  XXXI.     PROB. 

To  draw  a  straight  line  through  a  given  point  parallel  to  a  given  straight 

line. 

Let  A  be  the  given  point,  and  BC  the  given  straight  line,  it  is  required 
to  draw  a  straight  line  through  the  point  A,  parallel  to  the  straight  line 
BC. 


30  ELEMENTS 

In  BC  take  any  point  D,  and  join  —,- 

AD  ;    and  at  the   point  A,  in  the  ■" 

straight  line  AD,  make  (23.  1.)  the 
angle  DAE  equal  to  the  angle  ADC  ; 


and  produce  the  straight  line  E  A  to  F.  D  JJ  C 

Because  the  straight  line  AD,  which  meets  the  two  straight  lines  BC, 
EF,  makes  the  alternate  angles  EAD,  ADC  equal  to  one  another,  EF  is 
parallel  (27.  1.)  to  BC.  Therefore  the  straight  line  EAF  is  drawn 
through  the  given  point  A  parallel  to  the  given  straight  line  BC. 

PROP.  XXXII.     THEOR. 

If  a  side  of  any  triangle  be  produced,  the  exterior  angle  is  equal  to  the  two 
interior  and  opposite  angles ;  and  the  three  interior  angles  of  every  triangle 
are  equal  to  two  right  angles. 

Let  ABC  be  a  triangle,  and  let  one  of  its  sides  BC  be  produced  to  D  ; 
the  exterior  angle  ACD  is  equal  to  the  two  interior  and  opposite  angles 
CAB,  ABC  ;  and  the  three  interior  angles  of  the  triangle,  viz.  ABC,  BCA, 
CAB,  are  together  equal  to  two  right  angles. 

Through  the  point  C  draw 
CE  parallel  (31.  1.)  to  the 
straight  line  AB  ;  and  because 
AB  is  parallel  to  CE,  and  AC 
meets  them,  the  alternate  an- 
gles BAC,  ACE  are  equal  (29. 
r.)  Again,  because  AB  is  pa- 
rallel to  CE,  and  BD  falls  upon 
them,  the  exterior  angle  ECD  is  equal  to  the  interior  and  opposite  angle 
ABC,  but  the  angle  ACE  was  shewn  to  be  equal  to  the  angle  BAC  ; 
therefore  the  whole  exterior  angle  ACD  is  equal  to  the  two  interior  and 
opposite  angles  CAB,  ABC ;  to  these  angles  add  the  angle  ACB,  and 
the  angles  ACD,  ACB  are  equal  to  the  three  angles  CBA,  BAC,  ACB  ; 
but  the  angles  ACD,  ACB  are  equal  (13.  1.)  to  two  right  angles ;  there- 
fore also  the  angles  CBA,  BAC,  ACB  are  equal  to  two  right  angles. 

Cor.  1.  All  the  interior  angles  of  any  rectilineal  figure  are  equal  to 
twice  as  many  right  angles  as  the  figure  has  sides,  wanting  four  right  angles. 

For  any  rectilineal  figure  ABCDE  can  be  divided  into  as  many  trian- 
gles as  the  figure  has  sides,  by  drawing  straight  lines  from  a  point  F 
within  the  figure  to  each  of  its  angles.  And,  by  the  preceding  proposition, 
ail  the  angles  of  these  triangles  are  equal 
to  twice  as  many  right  angles  as  there 
are  triangles,  that  is,  as  there  are  sides 
of  the  figure  ;  and  the  same  angles  are 
equal  to  the  angles  of  the  figure,  together 
with  the  angles  at  the  point  F,  which 
is  the  common  vertex  of  the  triangles ; 
that  is,  (2  Cor.  15.  1.)  together  with  four 
right  angles.  Therefore,  twice  as  many 
right  angles  as  the  figure  has  sides,  are 
equal  to  all  the  angles  of  the  figure,  to- 


OF  GEOMETRY.    BOOK  I.  31 

gether  with  four  right  angles,  that  is,  the  angles  of  the  figure  are  equal 
to  twice  as  many  right  angles  as  the  figure  has  sides,  wanting  four. 

Cor.  2.  All  the  exterior  angles  of  any  rectilineal  figure  are  together 
equal  to  four  right  angles. 

Because  every  interior  angle 

ABC,  with  its  adjacent  exterior 

ABD,  is  equal  (13.  1.)  to  two 
right  angles  ;  therefore  all  the 
interior,  together  with  all  the 
exterior  angles  of  the  figure, 
are  equal  to  twice  as  many 
right  angles  as  there  are  sides 
of  the  figure ;  that  is,  by  the 
foregoing  corollary,  they  are 
equal  to  all  the  interior  angles 
of  the  figure,  together  with 
four  right  angles  ;  therefore  all 
the  exterior  angles  are  equal  to  four  right  angles. 

Cor.  3.  Two  angles  of  a  triangle  being  given,  or  merely  their  sum,  tho 
third  will  be  found  by  subtracting  that  sum  from  two  right  angles. 

CoR.  4.  If  two  angles  of  one  triangle  are  respectively  equal  to  two  an- 
gles of  another,  the  third  angles  will  also  be  equal,  and  the  two  triangles 
will  be  mutually  equiangular. 

Cor.  5.  In  any  triangle  there  can  be  but  one  right  angle  ;  for  if  there 
were  two,  the  third  angle  must  be  nothing.  Still  less  can  a  triangle  have 
more  than  one  obtuse  angle. 

Cor.  6.  In  every  right-angled  triangle,  the  sum  of  the  two  acute  an- 
gles is  equal  to  one  right  angle. 

Cor.  7.  Since 'every  equilateral  triangle  (Cor.  5.  1.)  is  also  equian- 
gular, each  of  its  angles  will  be  equal  to  the  third  part  of  two  right  angles  ; 
so  that  if  the  right  angle  is  expressed  by  unity,  the  angle  of  an  equilateral 
triangle  will  be  expressed  by  f  of  one  right  angip. 

CoR.  8.  The  sum  of  the  angles  in  a  quadrilateral  is  equal  to  two  right 
angles  multiplied  by  4  —  2,  which  amounts  to  four  right  angles  ;  hence,  if 
all  the  angles  of  a  quadrilateral  are  equal,  each  of  them  will  be  a  right  an- 
gle ;  a  conclusion  which  sanctions  the  Definitions  25  and  26,  where  the 
four  angles  of  a  quadrilateral  are  said  to  be  right,  in  the  case  of  the  rectan- 
gle and  the  square. 

Cor.  9.  The  sum  of  the  angles  of  a  pentagon  is  equal  to  two  right  an- 
gles multiplied  by  5  —  2,  which  amounts  to  six  right  angles  ;  hence,  when 
a  pentagon  is  equiangular,  each  angle  is  equal  to  the  fifth  part  of  six  right 
angles,  or  |  of  one  right  angle. 

Cor.  10.  The  sum  of  the  angles  of  a  hexagon  is  equal  to  2  x  (6  —  2), 
or  eight  right  angles ;  hence,  in  the  equiangular  hexagon,  each  angle  is 
the  sixth  oart  of  eight  right  angles,  or  ^  of  one  right  angle. 

SCHOLIUM. 

When  (Cor.  1.)  is  applied  to  polygons,  which  have  re-entrant  angles, 
as  ABC  each  re-entrant  angle  must  be  regarded  as  greater  than  two  right 
angles. 


32 


ELEMENTS 


And,  by  joining  BD,  BE,  BF,  the 
figure  is  divided  into  four  triangles, 
which  contain  eight  right  angles  ; 
that  is,  as  many  times  two  right  an- 
gles as  there  are  units  in  the  number 
of  sides  diminished  by  two. 

But  to  avoid  all  ambiguity,  we  shall 
henceforth  limit  our  reasoning  to 
polygons  with  salient  angles,  which 
might  otherwise  be  named  convex 
polygons.  Every  convex  polygon  is 
such  that  a  straight  line,  drawn  at 
pleasure,  cannot  meet  the  contour  of 
the  polygon  in  more  than  two  points. 


PROP.  XXXIII.    THEOR. 

The  straight  lines  which  join  the  extremities  of  two  equal  and  parallel  straight 
lines,  towards  the  same  parts,  are  also  themselves  equal  and  parallel. 

Let  AB,  CD,  be  equal  and  parallel  straight  lines,  and  joined  towards 
the  same  parts  by  the  straight  lines  AC,  BD;  AC,  BD  are  also  equal  and 
parallel. 

Join  BC ;  and  because  AB  is  parallel 
to  CD,  and  BC  meets  them,  the  alternate 
angles  ABC,  BCD  are  equal  (29.  1.) ;  and 
because  AB  is  equal  to  CD,  and  BC  com- 
mon to  the  two  triangles  ABC,  DCB,  the 
two  sides  AB,  BC  are  equal  to  the  two 
DC,  CB  ;  and  the  angle  ABC  is  equal  to  C  "D 

the  angle  BCD;  therefore  the  base  AC  is  equal  (4.  1.)  to  the  base  BD, 
and  the  triangle  ABC  to  the  triangle  BCD,  and  the  other  angles  to  the 
other  angles  (4.  1.)  each  to  each,  to  which  the  equal  sides  are  opposite  ; 
therefore  the  angle  ACB  is  equal  to  the  angle  CBD  ;  and  because  the 
straight  line  BC  meets  the  two  straight  lines  AC,  BD,  and  makes  the  al- 
ternate angles  ACB,  CBD  equal  to  one  another,  AC  is  parallel  (27.  1.)  to 
BD  ;  and  it  was  shewn  to  be  equal  to  it. 

CoR.  1.  Hence,  if  two  opposite  sides  of  a  quadrilateral  are  equal  and 
parallel,  the  remaining  sides  will  also  be  equal  and  parallel,  and  the  figure 
will  be  a  parallelogram. 

Cor.  2.  And  every  quadrilateral,  whose  opposite  sides  are  equal,  is  a 
parallelogram,  or  has  its  opposite  sides  parallel. 

For,  having  drawn  the  diagonal  BC ;  then,  the  triangles  ABC,  CBD, 
being  mutually  equilateral  (hyp.),  they  are  also  mutually  equiangular 
(Th.  8.),  or  have  their  corresponding  angles  equal ;  consequently,  the  op- 
posite sides  are  parallel ;  namely,  the  side  AB  parallel  to  CD,  and  BD  pa- 
rallel to  AC  ;  and,  therefore,  the  figure  is  a  parallelogram. 

Cor.  3.  Hence,  also,  if  the  opposite  angles  of  a  quadrilateral  be  equal, 
the  opposite  sides  will  likewise  be  equal  and  parallel. 

For  all  the  angles  of  the  figure  being  equal  to  four  right  angles  (Cor.  8. 


OF  GEOMETRY.     BOOK  I.  33 

Th.  32.),  and  the  opposite  angles  being  mutually  equal,  each  pair  of  adja- 
cent angles  must  be  equal  to  two  right  angles  ;  therefore,  the  opposite  sides 
must  be  equal  and  parallel. 

PROP.  XXXIV.     THEOR. 

The  opposite  sides  and  angles  of  a  parallelogram  are  equal  to  one  another,  and 
the  diagonal  bisects  it ;  that  is,  divides  it  into^two  equal  parts. 

N.  B.     A  Parallelogram  is  a  four-sided  figure,  of  which  the  opposite  sides  are  parallel ;  and 
the  diameter  is  a  straight  line  joining  two  of  its  opposite  angles. 

Let  ACDB  be  a  parallelogram,  of  which  BC  is  a  diameter ;  the  oppo- 
site sides  and  angles  of  the  figure  are  equal  to  one  another  ;  and  the  diam- 
eter BC  bisects  it. 

Because  AB  is  parallel  to  CD,  and  BC 
meets  them,  the  alternate  angles  ABC, 
BCD  are  equal  (29. 1.)  to  one  another ;  and 
because  AC  is  parallel  to  BD,  and  BC  meets 
them,  the  alternate  angles  ACB,  CBD  are 
equal  (29.  1.)  to  one  another;  wherefore 
the  two  triangles  ABC,  CBD  have  two  an- 
gles ABC,  BCA  in  one,  equal  to  two  angles 

BCD,  CBD  in  the  other,  each  to  each,  and  the  side  BC,  which  is  adja- 
cent to  these  equal  angles,  common  to  the  two  triangles ;  therefore  their 
other  sides  are  equal,  each  to  each,  and  the  third  angle  of  the  one  to  the 
third  angle  of  the  other  (26.  1.)  ;  viz.  the  side  AB  to  the  side  CD,  and 
AC  to  BD,  and  the  angle  BAC  equal  to  the  angle  BDC.  And  because 
the  angle  ABC  is  equal  to  the  angle  BCD,  and  the  angle  CBD  to  the 
angle  ACB,  the  whole  angle  ABD  is  equal  to  the  whole  angle  ACD  : 
And  the  angle  BAC  has  been  shewn  to  be  equal  to  the  angle  BDC  :  there- 
fore the  opposite  sides  and  angles  of  a  parallelogram  are  equal  to  one  an- 
other ;  also,  its  diameter  bisects  it ;  for  AB  being  equal  to  CD,  and  BC 
common,  the  two  AB,  BC  are  equal  to  the  two  DC,  CB,  each  to  each  ; 
now  the  angle  ABC  is  equal  to  the  angle  BCD  ;  therefore  the  triangle 
ABC  is  equal  (4.  1.)  to  the  triangle  BCD,  and  the  diameter  BC  divides 
the  parallelogram  ACDB  into  two  equal  parts. 

CoR.  1.  Two  parallel  lines,  included  between  two  other  parallels,  are 
equal. 

Cor.  2.     Hence,  two  parallels  are  every  where  equally  distant. 

Cor.  3.  Hence,  also,  the  sum  of  any  two  adjacent  angles  of  a  paral- 
lelogram is  equal  to  two  right  angles. 

PROP.  XXXV.     THEOR. 

Parallelograms  upon  the  same  base  and  between  the  same  parallels,  are  equal 

to  one  another. 

(see  the  2d  AND  3d  figures.) 

Let  the  parallelograms  ABCD,  EBCF  be  upon  the  same  base  BC,  and 
between  the  same  parallels  AF,  BC  ;  the  parallelogram  ABCD  is  equal  to 
the  parallelogram  EBCF. 

5 


34 


ELEMENTS 


If  the  sides  AD,  DF  of  the  parallelo- 
grams ABCD,  DBCF  opposite  to  the  base 
BC  be  terminated  in  the  same  point  D  ; 
it  is  plain  that  each  of  the  parallelograms 
is  double  (34.  1.)  of  the  triangle  BDC  ; 
and  they  are  therefore  equal  to  one  an- 
other. 

But,  if  the  sides  AD,  EF,  opposite  to  the  base  BC  of  the  parallelograms 
ABCD,EBCF,  be  not  terminated  in  the  same  point ;  then,  because  ABCD 
is  a  parallelogram,  AD  is  equal  (34.  l.)to  BC  ;  for  the  same  reason  EF 
is  equal  to  BC  ;  wherefore  AD  is  equal  (1.  Ax.)  to  EF  ;  and  DE  is  com- 
mon ;  therefore  the  whole,  or  the  remainder,  AE  is  equal  (2.  or  3.  Ax.)  to 
the  whole,  or  the  remainder  DF  ;  now  AB  is  also  equal  to  DC  ;  therefore 
the  two  E  A,  AB  are  equal  to  the  two  FD,  DC,  each  to  each  ;  but  the  ex- 

A. D_E  FA        E  T>         F 


terior  angle  FDC  is  equal  (29. 1.)  to  the  interior  EAB,  wherefore  the  base 
EB  is  equal  to  the  base  FC,  and  the  triangle  EAB  (4.  1.)  to  the  triangle 
FDC.  Take  the  triangle  FDC  from  the  trapezium  ABCF,  and  from  the 
same  trapezium  take  the  triangle  EAB  ;  the  remainders  will  then  be  equal 
(3.  Ax. )  that  is,  the  parallelogram  AB  C  D  is  equal  to  the  parallelogram  E  B  CF. 

PROP.  XXXVI.     THEOR. 

Parallelograms  upon  equal  bases,  and  between  the  same  parallels^  are  equal  to 

one  another. 

Let  ABCD,  EFGH  be  parallelograms  upon  equal  bases  BC,  FG,  and 
between  the  same  parallels  AH,        -  t%     T  TX 

BG  ;  the  parallelogram   ABCD      A  D    E  H 

is  equal  to  EFGH. 

Join  BE,  CH  ;  and  because 
BC  is  equal  to  FG,  and  FG  to 
(34.1.)  EH,  BC  is  equal  to  EH; 
and  they  are  parallels,  and  join- 
ed towards  the  same  parts  by  the 
straight  lines  BE,  CH  :  But 
straight  lines  which  join  equal  and  parallel  straight  lines  towards  the  same 
parts,  are  themselves  equal  and  parallel  (33.  1.)  ;  therefore  EB,  CH  are 
both  equal  and  parallel,  and  EBCH  is  a  parallelogram  ;  and  it  is  equal 
(35.  1.)  to  ABCD,  because  it  is  upon  the  same  base  BC,  and  between  the 
same  parallels  BC,  AH  :  For  the  like  reason,  the  parallelogram  EFGH 
is  equal  to  the  same  EBCH  :  Therefore  also  the  parallelogram  ABCD  is 
equal  to  EFGH. 


OF  GEOMETRY.    BOOK  I.  $  35 


PROP.  XXXVII.     THEOR. 

Triangles  upon  the  same  base,  and  hetioeen  the  same  parallels^  are  equal  to  one 

another. 

Let  the  triangles  ABC,  DBG  be  upon  the  same  base  BG,  and  between 
the  same  parallels,  AD,  BG  :    The 
triangle  ABC  is  equal  to  the  trian- 
gle DBG. 

Produce  AD  both  ways  to  the 
points  E,  F,  and  through  B  draw  (31. 
1.)  BE  parallel  to  GA  ;  and  through 
G  draw  CF  parallel  to  BD  :  There- 
fore, each  of  the  figures  EBGA, 
DBCF  is  a  parallelogram;  andEBCA 
is  equal  (35.  1.)  to  DBCF,  because  they  are  upon  the  same  base  BG,  and 
between  the  same  parallels  BG,  EF  ;  but  the  triangle  ABC  is  the  half  of 
the  parallelogram  EBGA,  because  the  diameter  AB  bisects  (34.  1.)  it; 
and  the  triangle  DBG  is  the  half  of  the  parallelogram  DBCF,  because 
the  diameter  DC  bisects  it ;  and  the  halves  of  equal  things  are  equal  (7. 
Ax.) ;  therefore  the  triangle  ABC  is  equal  to  the  triangle  DBG. 

PROP.  XXXVIII.    THEOR. 

Triangles  upon  equal  bases,  and  between  the  same  parallels,  are  equal  to  one 

another. 

Let  the  triangles  ABC,  DEF  be  upon  equal  bases  BG,  EF,  and  between 
the  same  parallels  BF,  AD  :  The  triangle  ABC  is  equal  to  the  triangle  DEF. 

Produce  AD  both  ways  to  the  points  G,  H,  and  through  B  draw  BG 
parallel  (31.  1.)  to  GA,  and  through  F  draw  FH  parallel  to  ED  :  Then 
each  of  the  figures   GBGA,     ^  A  n  TT 

DEFH   is  a   parallelogram;     Ir  ^  -Lr Ji 

and  they  are  equal  to  (36.  1.) 
one  another,  because  they  aie 
upon  equal  bases  BG,  EF,  and 
between  the  same  parallels 
BF,   GH  ;    and  the  triangle 

ABC  is  the  half  (34.  1.)  of  the       ^ " — ±r-± ^ 

parallelogram  GBGA, because       ^  C    Jii  r 

the  diameter  AB  bisects  it;  and  the  triangle  DEF  is  the  half (34.  L)  of 
the  parallelogram  DEFH,  because  the  diameter  DF  bisects  it :  But  the 
halves  of  equal  things  are  equal  (7.  Ax.) ;  therefore  the  triangle  ABC  is 
equal  to  the  triangle  DEF. 

PROP.  XXXIX.     THEOR. 

Equal  triangles  upon  the  same  base,  and  upon  the  same  side  of  it,  are  between 

the  same  parallels. 

Let  the  equal  triangles  ABC,  DBG  be  upon  the  same  base  BG,  and  upon 
the  same  side  of  it  ;  they  are  between  the  same  parallels. 


36  t  ELEMENTS 

Join  AD  ;  AD  is  parallel  to  BC  ;  for,  if  it  is  not,  through  the  point  A 
draw  (31.  1.)  AE  parallel  to  BC,  and  join  EC  : 
The  triangle  ABC,  is  equal  (37,  1.)  to  the  tri- 
angle EBC,  because  it  is  upon  the  same  base 
BC,  and  between  the  same  parallels  BC,  AE  : 
But  the  triangle  ABC  is  equal  to  the  triangle 
BDC  ;  therefore  also  the  triangle  BDC  is  equal 
to  the  triangle  EBC,  the  greater  to  the  less, 
which  is  impossible  :  Therefore  AE  is  not  par- 
allel to  BC.  In  the  same  manner,  it  may  be 
demonstrated  that  no  other  line  but  AD  is  parallel  to  BC  ;  AD  is  there- 
fore parallel  to  it. 

PROP.  XL.    THEOR. 

Equal  triangles  on  the  same  side  of  bases  which  are  equal  and  in  the  same 
straight  line^  are  between  the  same  parallels. 

Let  the  equal  triangles  ABC,  DEF  be  upon  equal  bases  BC,  EF,  in 
the  same  straight  line  BF,  and  to- 
wards the  same  parts  ;  they  are  be- 
tween the  same  parallels. 

Join  AD  ;  AD  is  parallel  to  BC  ; 
for,  if  it  is  not,  through  A  draw  (31. 
L)  AG  parallel  to  BF,  and  join  OF. 
The  triangle  ABC  is  equal  (38.  1.) 
to  the  triangle  GEF,  because  they 
are  upon  equal  bases  BC,  EF,  and 
between  the  same  parallels  BF, 
AG :  But  the  triangle  ABC  is  equal  to  the  triangle  DEF  ;  therefore  also 
the  triangle  DEF  is  equal  to  the  triangle  GEF,  the  greater  to  the  less, 
which  is  impossible  ;  therefore  AG  is  not  parallel  to  BF  ;  and  in  the  same 
manner  it  may  be  demonstrated  that  there  is  no  other  parallel  to  it  but 
AD ;  AD  is  therefore  parallel  to  BF. 

PROP.  XLL    THEOR. 

If  a  parallelogram  and  a  triangle  he  upon  the  same  hase^  and  between  the 
same  parallel ;  the  parallelogram  is  double  of  the  triangle.   • 

Let  the  parallelogram  ABCD  and  the  tri- 
angle EBC  be  upon  the  same  base  BC  and 
between  the  same  parallels  BC,  AE ;  the 
parallelogram  ABCD  is  double  of  the  trian- 
gle EBC. 

Join  AC  ;  then  the  triangle  ABC  is  equal 
(37.  1.)  to  the  triangle  EBC,  because  they 
are  upon  the  same  base  BC,  and  between  the 
same  parallels  BC,  AE.  But  the  parallelo- 
gram ABCD  is  double  (34.  1.)  of  the  triangle  B 
ABC,  because  the  diameter  AC  divides  it 
into  two  equal  parts ;  wherefore  ABCD  is  also  double  of  the  triangle  EBC. 


OF  GEOMETRY.    BOOK  I. 


37 


PROP.  XLII.     PROB. 

To  describe  a  parallelogram  that  shall  he  equal  to  a  given  triangle,  and  have 
one  of  its  angles  equal  to  a  given  rectilineal  angle. 

Let  ABC  be  the  given  triangle,  and  D  the  given  rectilineal  angle.  It 
is  required  to  describe  a  parallelogram  that  shall  be  equal  to  the  given  tri- 
angle ABC,  and  have  one  of  its  angles  equal  to  D. 

Bisect  (10.  1.)  BC  in  E,  join  AE,  and  at  the  point  E  in  the  str^iight  line 
EC  make  (23.  1.)  the  angle  CEF  equal  to  D  ;  and  through  A  draw  (31. 
1.)  AG  parallel  to  BC,  and  through  C  draw  CG  (31.  1.)  parallel  to  EF ; 
Therefore  FECG  is  a  parallelogram  :  a 
And  because  BE  is  equal  to  EC,  the 
triangle  ABE  is  likewise  equal  (38. 
1.)  to  the  triangle  AEC,  since  they 
are  upon  equal  bases  BE,  EC,  and 
between  the  same  parallels  BC,AG; 
therefore  the  triangle  ABC  is  double 
jf  the  triangle  AEC.  And  the  paral- 
lelogram FECG  is  likewise  double 
(41.  1.)  of  the  triangle  AEC,  because 
it  is  upon  the  same  base,  and  between 
the  same  parallels :  Therefore  the  parallelogram  FECG  is  equal  to  the 
triangle  ABC,  and  it  has  one  of  its  angles  CEF  equal  to  the  given  angle 
D :  Wherefore  there  has  been  described  a  parallelogram  FECG  equal  to 
a  given  triangle  ABC,  having  one  of  its  angles  CEF  equal  to  the  given 
angle  D. 

Cor.  Hence,  if  the  angle  D  be  a  right  angle,  the  parallelogram  EFGC 
will  be  a  rectangle,  equivalent  to  the  triangle  ABC  ;  and  therefore  the 
same  construction  will  apply  to  the  problem :  to  make  a  triangle  equivalent 
to  a  given  rectangle. 

PROP.  XLIII.     THEOR. 

The  complements  of  the  parallelograms  which  are  about  the  diameter  of  any 
parallelogram,  are  equal  to  one  another. 

Let  ABCD  be  a  parallelogram  of  which  the  diameter  is  AC  ;  let  EH, 
FG  be  the  parallelograms  about  AC,  that  is,  through  which  AC  passes,  and 
let  BK,  KD  be  the  other  parallelograms, 
which  make  up  the  whole  figure  ABCD, 
and  are  therefore  called  the  complements  ; 
The  complement  BK  is  equal  to  the  com- 
plement KD. 

Because  ABCD  is  a  parallelogram  and 
AC  its  diameter,  the  triangle  ABC  is 
equal  (34.  1.)  to  the  triangle  ADC  :  And 
because  EKHA  is  a  parallelogram,  and 
AK  its  diameter,  the  triangle  AEK  is 
equal  to  the  triangle  AHK  :  For  the  same 
reason,  the  triangle  KGC  is  equal  to  the 


38 


ELEMENTS 


triangle  KFC.  Then,  because  the  triangle  AEK  is  equal  to  the  triangle 
AHK,  and  the  triangle  KGC  to  the  triangle  KFC  ;  the  triangle  AEK,  to- 
gether with  the  triangle  KGC,  is  equal  to  the  triangle  xVHK,  together  with 
the  triangle  KFC  :  But  the  whole  triangle  ABC  is  equal  to  the  whole 
ADC  ;  therefore  the  remaining  complement  BK  is  equal  to  the  remaining 
complement  KD. 


PROP.  XLIV.     PROB. 

To  a  given  straight  line  to  apply  a  parallelogram,  which  shall  he  equal  to  a  given 
triangle,  and  have  one  of  its  angles  ejual  to  a  given  rectilineal  angle. 

Let  AB  be  the  given  straight  line,  and  C  the  given  triangle,  and  D  the 
given  rectilineal  angle.  It  is  required  to  apply  to  the  straight  line  AB  a 
parallelogram  equal  to  the  triangle  C,  and  having  an  angle  equal  to  D. 
Make  (42.  1.)  the  parallelogram  BEFG  equal  to  the  triangle  C,  having  the 


angle  EBG  equal  to  the  angle  D,  and  the  side  BE  in  the  same  straight 
line  with  AB  :  produce  FG  to  H,  and  through  A  draw  (31.  1.)  AH  parallel 
to  BG  or  EF,  and  join  HB.  Then  because  the  straight  line  HF  falls  upon 
the  parallels  AH,  EF,  the  angles  AHF,  HFE,  are  together  equal  (29.  1.) 
to  two  right  angles  ;  wherefore  the  angles  BHF,  HFE  are  less  than  two 
right  angles  ;  But  straight  lines  which  with  another  straight  line  make  the 
interior  angles,  upon  the  same  side  less  than  two  right  angles,  do  meet  if  pro- 
duced (1  Cor.  29. 1.) :  Therefore  HB,  FE  will  meet,  if  produced  ;  let  them 
meet  in  K,  and  through  K  draw  KL  parallel  to  E A  or  FH,  and  produce  HA, 
GB  to  the  points  L,  M  :  Then  HLKF  is  a  parallelogram,  of  which  the  diam- 
eter is  HK,  and  AG,  ME  are  the  parallelograms  about  HK ;  and  LB,  BF  are 
the  complements ;  therefore  LB  is  equal  (43.  I.)  to  BF  :  but  BF  is  equal 
to  the  triangle  C  ;  wherefore  LB  is  equal  to  the  triangle  C  ;  and  because 
the  angle  GBE  is  equal  (15.  1.)  to  the  angle  ABM,  and  likewise  to  the  an- 
gle D  ;  the  angle  ABM  is  equal  to  the  angle  D  :  Therefore  the  parallelo- 
gram LB,  which  is  applied  to  the  straight  line  AB,  is  equal  to  the  triangle 
C,  and  has  the  angle  ABM  equal  to  the  angle  D. 

CoR.  Hence,  a  triangle  may  be  converted  into  an  equivalent  rectangle^ 
having  a  side  of  a  given  length  :  for,  if  the  angle  D  be  a  right  angle,  and 
AB  the  given  side,  the  parallelogram  ABML  will  be  a  rectangle  equiva- 
lent to  the  triangle  C. 


OF  GEOMETRY.  BOOK  I.  39 


PROP.  XLV.  PROB. 

To  describe  a  parallelogram  equal  to  a  given  rectilineal  figure,  and  having 
an  angle  equal  to  a  given  rectilineal  angle. 

Let  ABCD  be  the  given  rectilineal  figure,  and  E  the  given  rectilineal 
angle.  It  is  required  to  describe  a  parallelogTam  equal  to  ABCD,  and  hav- 
ing an  angle  equal  to  E. 

Join  DB,  and  describe  (42.  1.)  the  parallelogram  FH  equal  to  the  tri- 
angle ADB,  and  having  the  angle  HKF  equal  to  the  angle  E  ;  and  to  the 
straight  line  GH  (44.  1.)  apply  the  parallelogram  GM  equal  to  the  triangle 
DBG,  having  the  angle  GHM  equal  to  the  angle  E.  And  because  the  an- 
gle E  is  equal  to  each  of  the  angles  FKH,  GHM,  the  angle  FKH  is  equal 
to  GHM  ;  add  to  each  of  these  the  angle  KHG  ;  therefore  the  angles 
FKH,  KHG  are  equal  to  the  angles  KHG,  GHM  ;  but  FKH,  KHG  are 
equal  (29.  1.)  to  two  right  angles ;  therefore  also  KHG,  GHM  are  equal 
to  two  right  angles  ;  and  because  at  the  Doint  H  in  the  straight  lines  GH, 


the  two  straight  lines  KH,  HM,  upon  the  opposite  sides  of  GH,  make  the 
adjacent  angles  equal  to  two  right  angles,  KH  is  in  the  same  straight  line 
(14.  1.)  with  HM.  And  because  the  straight  line  HG  meets  the  parallels 
KM,  FG,  the  alternate  angles  MHG,  HGF  are  equal  (29.  1.) ;  add  to  each 
of  these  the  angle  HGL  :  therefore  the  angles  MHG,  HGL,  are  equal  to 
the  angles  HGF,  HGL  :  But  the  angles  MHG,  HGL,  are  equal  (29.  l.)to 
two  right  angles  ;  wherefore  also  the  angles  HGF,  HGL,  are  equal  to  two 
right  angles,  and  FG  is  therefore  in  the  same  straight  line  with  GL.  And 
because  KF  is  parallel  to  HG,  and  HG  to  ML,  KF  is  parallel  (30.  L)  to 
ML  ;  but  KM,  FL  are  parallels  :  wherefore  KFLM  is  a  parallelogram. 
And  because  the  triangle  ABD  is  equal  to  the  parallelogram  HF,  and  the 
triangle  DBG  to  the  parallelogi-am  GM,  the  whole  rectilineal  figure  ABCD 
is  equal  to  the  whole  parallelogram  KFLM  ;  therefore  the  parallelogram 
KFLM  has  been  described  equal  to  the  given  rectilineal  figure  ABCD,  hav- 
ing the  angle  FKM  equal  to  the  given  angle  E. 

CoR.  From  this  it  is  manifest  how  to  a  given  straight  line  to  apply  a 
parallelogram,  which  shall  have  an  angle  equal  to  a  given  rectilineal  angle, 
and  shall  be  equal  to  a  given  rectilineal  figure,  viz.  by  applying  (44.  \.) 
to  the  given  straight  line  a  parallelogram  equal  to  the  first  triangle  ABD, 
and  having  an  angle  equal  to  the  given  angle. 


40  ELEMENTS 

PROP.  XLYI.     PROB. 
To  describe  a  square  upon  a  given  straight  line. 

Let  A  B  be  the  given  straight  line  :  it  is  required  to  describe  a  square 
upon  AB. 

From  the  point  A  draw  (IL  1.)  AC  at  right  angles  to  AB  ;  and  make 
(3.  1.)  AD  equal  to  AB,  and  through  the  point  D  draw  DE  parallel  (31.1.) 
to  AB,  and  through  B  draw  BE  parallel  to  AD  ;  therefore  ADEB  is  a  par- 
allelogram ;  whence  AB  is  equal  (34.  1.)  to  DE,  and  AD  to  BE  ;  but  BA 
is  equal  to  AD  :  therefore  the  four  straight  -^ 
lines  BA,  AD,  DE,  EB  are  equal  to  one  an-  ' 
other,  and  the  parallelogram  ADEB  is  equi- 
lateral ;    it  is    likewise  rectangular ;    for  the 

straight  line  AD  meeting  the  parallels,  AB,  DE,      D  | • — {E 

makes  the  angles  BAD,  ADE  equal  (29.  1.)  to 
two  right  angles  ;  but  BAD  is  a  right  angle  ; 
therefore  also  ADE  is  a  right  angle  now  the 
opposite  angles  of  parallelograms  are  equal  (34. 
1.) ;  therefore  each  ofthe  opposite  angles  ABE, 
BED  is  a  right  angle ;  wherefore  the  figure 

ADEB  is  rectangular,  and  it  has  been  demon-       . , I 

strated  that  it  is  equilateral ;  it  is  therefore  a      "**•  o 

square,  and  it  is  described  upon  the  given  straight  line  AB, 

Cor.  Hence  every  parallelogram  that  has  one  right  angle  has  all  its  an- 
gles right  angles. 

PROP.  XLVIL     THEOR. 

In  any  right  angled  triangle,  the  square  which  is  described  upon  the  side 
subtending  the  right  angle,  is  equal  to  the  squares  described  upon  the  sides 
which  contain  the  right  angle. 

Let  ABC  be  a  right  angled  triangle  having  the  right  angle  BAC ;  the 
square  described  upon  the  side  BC  is  equal  to  the  squares  described  upon 
BA,  AC. 

On  BC  describe  (46.  1.)  the  square  BDEC,  and  on  BA,  AC  the  squares 
GB,  HC  ;  and  through  A  draw  (31.  1.)  AL  parallel  to  BD  or  CE,  and  join 
AD,  FC  ;  then,  because  each  of  the  angles  BAC,  BAG  is  a  right  angle 
(25.  dcf.),  the  two  straight  lines  AC,  AG  upon  the  opposite  sides  of  AB, 
make  with  it  at  the  point  A  the  adjacent  angles  equal  to  two  right  an- 
gles;  therefore  CA  is  in  the.  same  straight  line  (14.  1.)  with  AG;  for 
the  same  reason,  AB  and  AH  are  in  the  same  straight  line.  Now  be- 
cause the  angle  DBC  is  equal  to  the  angle  FBA,  each  of  them  being  a 
right  angle,  adding  to  each  the  angle  ABC,  the  whole  angle  DBA  will  be 
equal  (2.  Ax.)  to  the  whole  FBC  ;  and  because  the  two  sides  AB,  BD, 
are  equal  to  the  two  FB,  BC  each  to  each,  and  the  angle  DBA  equal  to 
the  angle  FBC,  therefore  the  base  AD  is  equal  (4.  1.)  to  the  base  FC, 
and  the  triangle  ABD  to  the  triangle  FBC.  But  the  parallelogram  BL 
is  double  (41.  1.)  of  the  triangle  ABD,  because  they  are  upon  the  same 
base,  BD,  and  between  the  same  parallels,  BD,  AL;  and  the  square  GB 


OF  GEOMETRY.    BOOK  I. 


41 


is  double  of  the  triangle  BFC  be- 
cause these  also  are  upon  the  same 
base  FB,  and  between  the  same  par- 
allels FB,  GC.  Now  the  doubles 
of  equals  are  equal  (6.  Ax.)  to  one  an- 
other ;  therefore  the  parallelogram 
BL  is  equal  to  the  square  GB  :  And 
in  the  same  manner,  by  joining  AE, 
BK,  it  is  demonstrated  that  the  par- 
allelogram CL  is  equal  to  the  square 
HC.  Therefore,  the  whole  square 
BDEC  is  equal  to  the  two  sqi^ares 
GB,  HC  ;  and  the  square  BDEC  is 
described  upon  the  straight  line  BG, 
and  the  squares  GB,  HC  upon  BA, 
AC  :  wherefore  the  square  upon  the 
side  BC  is  equal  to  the  squares  upon 
the  sides  BA,  AC. 

Cor.  1.  Hence,  the  square  of  one  of  the  sides  of  a  right  angled  triangle 
is  equivalent  to  the  square  of  the  hypotenuse  diminished  by  the  square  of 
the  other  side  ;  which  is  thus  expressed  :  AB^^rBC'* — AC^. 

Cor.  2.  If  AB  =  AC  ;  that  is,  if  the  triangle  ABC  be  right  angled  and 
isosceles;  BC2=2AB2=2AC2  ;  therefore,  BC  =  AB/ 2. 

Cor.  3.  Hence,  also,  if  two  right  angled  triangles  have  two  sides  of 
the  one,  equal  to  two  corresponding  sides  of  the  other ;  their  third  sides 
will  also  be  equal,  and  the  triangles  will  be  identical. 

PROP.  XLVHI.    THEOR. 

If  the  square  described  upon  one  of  the  sides  of  a  triangle^  he  equal  to  the 
squares  described  upon  the  other  txoo  sides  of  it ;  the  angle  contained  by 
these  two  sides  is  a  right  angle. 

If  the  square  described  upon  BC,  one  of  the  sides  of  the  triangle  ABC, 
be  equal  to  the  squares  upon  the  other  sides  BA,  AC,  the  angle  BAG  is 
a  right  angle. 

From  the  point  A  draw  (11.  1.)  AD  at  right  angles  to  AC,  and  make 
AD  equal  to  BA,  and  join  DC.  Then  because  DA  is  equal  to  AB,  the 
square  of  DA  is  equal  to  the  square  of  AB  ;  To 
each  of  these  add  the  square  of  AC  ;  therefore  the 
squares  of  DA,  AC  are  equal  to  the  squares  of  BA, 
AC.  But  the  square  of  DC  is  equal  (47.  1.)  to 
the  squares  of  DA,  AC,  because  DAC  is  a  right 
angle ;  and  the  square  of  BC,  by  hypothesis,  is 
equal  to  the  squares  of  BA,  AC  ;  therefore,  the 
square  of  DC  is  equal  to  the  square  of  BC ;  and 
therefore  also  the  side  DC  is  equal  to  the  side  BC. 
And  because  the  side  DA  is  equal  to  AB,  and  AC 
common  to  the  two  triangles  DAC,  BAG,  and  the  base  DC  likewise  equal 
to  the  base  BC,  the  angle  DAC  is  equal  (8.  1.)  to  the  angle  BAG  ;  But 
DAC  is  a  right  angle ;  therefore  also  BAG  is  a  right  angle. 

6 


42  ELEMENTS 

ADDITIONAL  PROPOSITIONS. 

PROP.  A.     THEOR. 

A  perpendicular  is  the  shortest  line  that  can  be  drawn  from  a  point,  situated 
without  a  straight  line,  to  that  line  :  any  two  oblique  lines  drawn  from  the 
same  point  on  different  sides  of  the  perpendicular, cuttijig  off  equal  distances 
on  the  other  line,  will  he  equal ;  and  any  two  other  oblique  lines,  cutting  off 
unequal  distances,  the  one  which  lies  farther  from  the  perpendicular  will 
be  the  longer. 

If  AB,  AC,  AD,  &c.  be  lines  drawn  from  the  given  point  A,  to  the  in- 
definite straight  line  DE,  of  which  AB  is  perpendicular;  then  shall  the 
perpendicular  AB  be  less  than  AC,  and  AC  less  than  AD,  and  so  on. 

For,  the  angle  ABC  being  a  right  one, 
the  angle  ACB  is  acute,  (17.  1.)  or  less 
than  the  angle  ABC.  But  the  less  angle 
of  a  triangle  is  subtended  by  the  less  side 
(19. 1.)  therefore,  the  side  AB  is  less  than 
the  side  AC. 

Again,  if  BC=:BE ;  then  the  two  ob- 
lique lines  AC,  AE,  are  equal.  For  the 
side  AB  is  common  to  the  two  triangles 
ABC,  ABE,  and  the  contained  angles  ABC 
and  ABE  equal;  the  two  triangles  must 
be  equal  (4.  1.) ;  hence  AE,  AC  are  equal. 

Finally,  the  angle  ACB  being  acute,  as  before,  the  adjacent  angle  ACD 
will  be  obtuse  ;  since  (13.  1.)  these  two  angles  are  together  equal  to  two 
right  angles;  and  the  angle  ADC  is  acute,  because  the  angle  ABD  is 
right ;  consequently,  the  angle  ACD  is  greater  than  the  angle  ADC  ;  and, 
since  the  greater  side  is  opposite  to  the  greater  angle  (19.  1.) ;  therefore 
the  side  AD  is  greater  than  the  side  AC. 

CoR.  1.  The  perpendicular  measures  the  true  distance  of  a  point  from 
a  line,  because  it  is  shorter  than  any  other  distance. 

Cor.  2.  Hence,  also,  every  point  in  a  perpendicular  at  the  middle  point 
of  a  given  straight  line,  is  equally  distant  from  the  extremities  of  that  line. 

CoR.  3.  From  the  same  point,  three  equal  straight  lines  cannot  be 
drawn  to  the  same  straight  line ;  for  if  there  could,  we  should  have  two 
equal  oblique  lines  on  the  same  side  of  the  perpendicular,  which  is  impos- 
sible. 

PROP.  B.     THEOR. 

When  the  hypotenuse  and  one  side  of  a  right  angled  triangle,  are  respective- 
ly equal  to  the  hypotenuse  and  one  side  of  another ;  the  two  right  angled 
triangles  are  equal. 

Suppose  the  hypotenuse  AC=DF,  and  the  side  AB=DE  ;  the  right 
angled  triangle  ABC  will  be  equal  to  the  right  angled  triangle  DEF. 


OF  GEOMETRY.     BOOK  I. 


43 


Their  equality  would  be  manifest,  if  tlie  third  sides  BC  and  EF  were 
equal.  If  possible,  suppose  that  those  sides  are  not  equal,  and  that  BC  is  the 
greater.  Take  BH=EF(3. 1.);  andjoin  AH.  Thetriangle  ABH  =  DEF; 
for  the  right  angles  B  and  E  are 
equal,  the  side  AB  =  DE,  and  BH 
=EF  ;  hence,  these  triangles  are 
equal  (4.  1.),  and  consequently 
AHrrDF.  Now  (hy  hyp.),  we 
have  DF=AC;  and  therefore, 
AH=AC.  But  by  the  last  prop- 
osition, the  oblique  line  AC  can- 
not be  equal  to  the  oblique  line 
AH,  which  lies  nearer  to  the  per- 
pendicular AB  ;  therefore  it  is 
impossible  that  BC  can  differ 
from  EF  ;  hence,  then,  the  trian- 
gles ABC  and  DEF  are  equal. 


PROP.  C.     THEOR. 


Two  angles  are  equal  if  their  sides  he  parallel j  each  to  eachj  and  lying  in  the 

same  direction. 


If  the  straight  lines  AB,  AC  be  parallel 
to  DF,  DE ;  the  angle  BAC  is  equal  to 
EDF. 

For,  draw  GAt)  through  the  vertices. 
And  since  AB  is  parallel  to  DF,  the  ex- 
terior angle  GAB  is  (29.  1.)  equal  to  GDF  ; 
and,  for  the  same  reason,  GAC  is  equal  to 
GDE  ;  there  consequently  remains  the  an- 
gle BAC^EDF. 


Cor.     If  BA,  AC  be  produced  to  I  and  H,  the  angle  BAC=HAI ; 
hence,  the  angle  HAI  is  also  equal  to  EDF. 


SCHOLIUM. 


The  restriction  of  this  proposition  to  the  case  where  the  side  AB  lies 
in  the  same  direction  with  DF,  and  AC  in  the  same  direction  with  DE, 
is  necessary ;  because  the  angle  CAI  would  have  its  sides  parallel  to  those 
of  the  angle  EDF,  but  would  not  be  equal  to  it.  In  that  case,  CAI  and 
EDF  would  be  together  equal  to  two  right  angles. 


44 


ELEMENTS 


PROP.  D.     PROB. 
Two  angles  of  a  triangle  being  given,  to  find  the  third. 

Draw  any  straight  line  CD  ;  at  a 
point  therein,  as  B,  make  the  angle 
CBA  equal  to  one  of  the  given  an- 
gles, and  the  angle  ABE  equal  to 
the  other :  the  remaining  angle  EBD 
will  be  the  third  angle  required  ;  be- 
cause those  three  angles  (Cor.  13.  1.) 
are  together  equal  to  two  right  angles. 


PROP.  E.     PROB. 
Two  angles  of  a  triangle  and  a  side  being  given,  to  construct  the  triangle. 

The  two  angles  will  either  be  both  adjacent  to  the  given  side,  or  the 
one  adjacent  and  the  other  opposite  :  in  the  latter  case,  find  the  third  angle 
(Prop.  D.) ;  and  the  two  adjacent  angles  will  thus  be  known. 

Draw  the  straight  line  BC  equal  to  the 
given  side  ;  at  the  point  B,  make  an  angle 
CBA  equal  to  one  of  the  adjacent  angles, 
and  at  C,  an  angle  BCA  equal  to  the  other ; 
the  two  lines  BA,  CA,  will  intersect  each 
other,  and  form  with  BC  the  triangle  re- 
quired ;  for  if  they  were  parallel,  the  an- 
gles B,  C,  would  be  together  equal  to  two 
right  angles,  and  therefore  could  not  be- 
long to  a  triangle  ;  hence,  BAC  will  be  the  triangle  required. 


PROP.  F.     PROB. 

« 

TtDO  sides  and  an  angle  opposite  to  one  of  them  being  given,  to  construct  the 

triangle. 

This  Problem  admits  of  two  cases. 

First.  When  the  given  angle 
is  obtuse,  make  the  angle  BCA 
equal  to  the  given  angle  ;  and  take 
C'A  equal  to  that  side  which  is 
adjacent  to  the  given  angle,  the 
arc  described  from  A  as  a  centre, 
with  a  radius  equal  to  AB,  the 
other  given  side,  would  cut  BC  on 
opposite  sides  of  C  ;  so  that  only      -o  p, 

one  obtuse  angled  triangle  could  be 
formed  ;  that  is,  the  triangle  BCA  will  be  the  triangle  required. 


OF  GEOMETRY.    BOOK  I.  45 

And,  if  the  given  angle  were  right,  although  two  triangles  would  be 
formed,  yet,  as  the  hypotenuse  would  meet  BO  at  equal  distances  from  the 
common  perpendicular,  these  triangles  would  be  equal. 

Secondly.  If  the  given  angle  be  acute,  and  the  side  opposite  to  it  greater 
than  the  adjacent  side,  the  same  mode  of  construction  will  apply  :  for,  mak- 
ing BCxi  equal  to  the  given  angle,  and  AC  equal  to  the  adjacent  side  ; 
then,  from  A  as  centre,  with  a  radius  equal  to  the  other  given  side,  describe 
an  arc,  cutting  CB  in  B  ;  draw  AB,  and  CAB  will  be  the  triangle  requi- 
red. 

But  if  the  given  angle  is  acute,  and  the  side  opposite  to  it  less  than  the 
other  given  side  ;  make  the  angle  CBA  equal  to  the  given  angle,  and  take 
BA  equal  to  the  adjacent  side  ;  then,  the  arc  described  from  the  centre  A, 
with  the  radius  AC  equal  to  the  opposite  side,  will  cut  the  straight  line 
BC  in  two  points  C^  and  C,  lying  on  the  same  side  of  B  :  hence,  there  will 
be  two  triangles  BAC^,  BAG,  either  of  which  will  satisfy  the  conditions 
of  the  problem. 

SCHOLIUM. 

In  the  last  case,  if  the  opposite  side  was  equal  to  the  perpendicular  from 
the  point  A  on  the  line  BC,  a  right  angled  triangle  would  be  formed.  And 
the  problem  would  be  impossible  in  all  cases,  if  the  opposite  side  was  less 
than  the  perpendicular  let  fall  from  the  point  A  on  the  straight  line  BC. 


PROP.  G.     PROB. 


To  find  a  triangle  that  shall  he  equivalent  to  any  given  rectilineal  figure. 

Let  ABODE  be  the  given  rectilineal  figure. 

Draw  the  diagonal  CE,  cutting  off  the  triangle  CDE  ;  draw  DF  paral- 
lel to  CE,  meeting  AE  produced,  and  join  CF ;  the  polygon  ABODE 
will  be  equivalent  to  the  polygon 
ABCF,  which  has  one  side  less  (J 

than  the  original  polygon. 

For  the  triangles  CDE,  CFE, 
have  the  base  CE  common,  and 
they  are  between  the  same  paral- 
lels ;  since  their  vertices  D,  F,  are 
situated  in  a  line  DF  parallel  to  the 
base  :  these  triangles  are  therefore 
equivalent  (37.  1.)  Draw,  now, 
the  diagonal  CA  and  BG  parallel 
to  it,  meeting  EA  produced  :  join 
CG,  and  the  polygon  ABCF  will  be 
reduced  to  an  equivalent  triangle  ; 
and  thus  the  pentagon  ABODE 
will  be  reduced  to  an  equivalent  triangle  GCF 


46  ELEMENTS 

The  same  process  may  be  applied  to  every  other  polygon  ;  for,  by  suc- 
cessively diminishing  the  number  of  its  sides,  one  being  retrenched  at  each 
step  of  the  process,  the  equivalent  triangle  will  at  length  be  found. 

CoR.  Since  a  triangle  may  be  converted  into  an  equivalent  rectangle, 
it  follows  that  any  polygon  may  he  reduced  to  an  equivalent  rectangle. 


PROP.  H.    PROB. 

To  find  the  side  of  a  square  that  shall  he  equivalent  to  the  sum  of  two  squares. 

Draw  the  two  indefinite  lines  AB,  AC,  per- 
pendicular to  each  other.  Take  AB  equal  to 
the  side  of  one  of  the  given  squares,  and  AC 
equal  to  the  other ;  join  BC  :  this  will  be  the 
side  of  the  square  required. 

For  the  triangle  BAC  being  right  angled, 
the  square  constructed  upon  BC  (47.  1.)  is 
equal  to  the  sum  of  the  squares  described  upon 
AB  and  AC. 

SCHOLIUM. 

A  square  may  be  thus  formed  that  shall  be  equivalent  to  the  sum  of  any 
number  of  squares  ;  for  a  similar  construction  which  reduces  two  of  them 
to  one,  will  reduce  three  of  them  to  two,  and  these  two  to  one,  and  so  of 
others. 

PROP.  L     PROB. 

To  find  the  side  of  a  square  equivalent  to  the  difference  of  two  given  squares. 

Draw,  as  in  the  last  problem,  (see  the  fig.)  the  lines  AC,  AD,  at  right  angles 
to  each  other,  making  AC  equal  to  the  side  of  the  less  square  ;  then,  from 
0  as  centre,  with  a  radius  equal  to  the  side  of  the  other  square,  describe 
an  arc  cutting  AD  in  D  :  the  square  described  upon  AD  will  be  equivalent 
to  the  difference  of  the  squares  constructed  upon  AC  and  CD. 

For  the  triangle  DAC  is  right  angled  ;  therefore,  the  square  described 
upon  DC  is  equivalent  to  the  squares  constructed  upon  AD  and  AC :  hence 
(Cor.  1.  47.  1.),  AD2=CD2_AC2. 

PROP.  K.     PROB. 

A  rectangle  being  given,  to  construct  an  equivalent  one,  having  a  side  oj  a 

given  length. 

Let  AEFH  be  the  given  rectangle,  and  produce  one  of  its  sides,  as  AH,  till 


OF  GEOMETRY.     BOOK  I. 


47 


HB  be  the  given  length,  and  draw  BFD 
meeting  the  prolongation  of  AE  in  D  ; 
then  produce  EF  till  FG  is  equal  to  HB  : 
draw  BGC,  HFK,  parallel  to  AED,  and 
through  the  point  D  draw  DKC  parallel 
to  AB  or  EG;  then,  the  rectangle 
GFKC,  having  the  side  FG  of  a  given 
length,  is  equal  to  the  given  rectangle 
AEFH  (43.  1.) 


Cor.     a  polygon  may  he  converted  into  an  equivalent  rectangle^  having  one 
of  its  sides  of  a  given  length. 


ELEMENTS 


OF 


GEOMETRY 


BOOK  U. 


DEFINITIONS. 


1.  Every  right  angled  parallelogram,  ox  rectangle,  is  said  to  be  contained 
by  any  two  of  the  straight  lines  which  are  about  one  of  the  right  an- 
gles. 

**  Thus  the  right  angled  paralltjlogram  AC  is  called  the  rectangle  contain- 
"  ed  by  AD  and  DC,  or  by  AD  and  AB,  &;c.  For  the  sake  of  brevity, 
"  instead  of  the  icctangle  contained  by  AD  and  DC,  we  shall  simply  say 
"  the  rectangle  AD  .  DC,  placing  a  point  between  the  two  sides  of  the 
"  rectangle." 

A.  In  Geometry,  the  product  of  two  lines  means  the  same  thing  as  their 
rectangle,  and  this  expression  has  passed  into  Arithmetic  and  Algebra, 
where  it  serves  to  designate  the  product  of  two  unequal  numbers  or 
quantities,  the  expression  square  being  employed  to  designate  the  pro- 
duct of  a  quantity  multiplied  by  itself. 
The  arithmetical  squares  of 

1,  2,  3,  &c.  are  1,  4,  9,  &c. 

So  likewise  the  square  de- 
scribed on  the  double  of 

a  line   is    evidently    four 

times  the  square  described 

on  a  single  one  ;  on  a  triple 

line  nine  times  that  on  a 

single  one,  &c. 


2.  In  every  parallelogram,  any  of  the 
parallelograms  about  a  diameter,  to- 
gether with  the  two  complements,  is 
called  a  Gnomon.  "  Thus  the  paral- 
"  lelogram  HG,  together  with  the 
"  complements  AF,  FC,  is  the  gno- 
"  mon  of  the  parallelogram  AC.  This 
*''  gnomon  may  also,  for  the  sake  of 
*  brevity,  be  called  the  gnomon  AGK 
•or  EHC.» 


OF  GEOMETRY.    BOOK  II. 


49 


PROP.  I.    THEOR. 

If  there  he  two  straight  lines,  one  of  which  is  divided  into  any  number  of 
parts ;  the  rectangle  contained  hy  the  two  straight  lines  is  equal  to  the 
rectangles  contained  by  the  undivided  line^  and  the  several  parts  of  the 
divided  line. 

Let  A  and  BC  be  two  straight  lines  ;  and  let  BC  be  divided  into  any 
parts  in  the  points  D,  E  ;  the  rectangle  A.BC  is  equal  to  the  several  rect- 
angles A.BD,  A.DE,  A.EC. 

From  the  point  B  draw  (Prop.  11.1.) 
BF  at  right  angles  to  BC,  and  make  BG        JJ  DEO 

equal  (Prop.  3.  1.)  to  A;  and  through 
G  draw  (Prop.  31.  1.)  GH  parallel  to 
BC  ;  and  through  D,  E,  C,  draw  DK, 
EL,  CH  parallel  to  BG ;  then  BH,  BK, 
DL,  and  EH  are  rectangles,  and  BH= 
BK-fDL+EH.  Gr 

But  BH  =  BG.BC=  A.BC,  because 
BG=A:  Also  BK  =  BG.BD=A.BD, 
because  BG=A;  and  DL=DK.DE=  P 
A.DE,  because  (34.  1.)  DK=BG=A. 
In  like  manner,  EH=A.EC.  Therefore  A.BC=A.BD+A.DE+A.EC  ; 
that  is,  the  rectangle  A.BC  is  equal  to  the  several  rectangles  A.BD,  A.DE, 
A.EC. 

SCHOLIUM. 

The  properties  of  the  sections  of  lines,  demonstrated  in  this  Book,  are 
easily  derived  from  Algebra.  In  this  proposition,  for  instance,  let  the  seg- 
ments of  BC  be  denoted  by  J,  c,  and  cZ;  then,  K(b-\-c-\-d):=kb-\- kc-\- kd. 


K    1 

.    H 

A 

PROP.  n.    THEOR. 

If  a  straight  line  he  divided  into  any  two  parts,  the  rectangles  contained  by  the 
whole  and  each  of  the  parts,  arc  together  equal  to  the  square  of  the  tuhole  line. 


c    B 


Let  the  straight  line  AB  be  divided  into  any 
two  parts  in  the  point  C  ;  the  rectangle  AB.BC3, 
together  with  the  rectangle  AB.AC,  is  equal  to 
the  square  of  AB  ;  or  AB.AC4-AB.BC=AB2. 

On  AB  describe  (Prop.  46.  1.)  the  square 
ADEB,  and  through  C  draw  CF  (Prop.  31.  1.) 
parallel  to  AD  or  BE  ;  then  AF+CE=AE. 
But  AF=:AD.AC=AB.AC,  because  AD=AB  ; 
CE=BE.BC=AB.BC;  and  AE=AB2.  There- 
fore AB.AC-|-AB.BC=AB2. 

SCHOLIUM. 

This  property  is  evident  from  Algebra  :  let  AB  be  denoted  by  a,  and  the 
segments  AC,  CB,  by  b  and  d,  respectively;  then,  a=b-{-d;  therefore, 
multiplying  both  members  of  this  equality  by  a,  we  shall  have  a-=:ab-{-ad 

7 


F    E 


50 


ELEMENTS 


/1r^ 


PROP.  III.     THEOR. 

If  a  straight  line  be  divided  into  any  two  parts,  the  rectangle  contained  by  the 
whole  and  one  of  the  parts,  is  equal  to  the  rectangle  contained  by  the  two 
parts,  together  with  the  square  of  the  aforesaid  part. 

Let  the  straight  line  AB  be  divided  into  two  parts,  in  the  point  C  ;  the 
rectangle  AB.BC  is  equal  to  the  rect- 
angle AC.BC,  together  with  BC^. 

Upon  BC  describe  (Prop.  46.  l.)the 
square  CDEB,  and  produce  ED  to  F, 
and  through  A  draw  (Prop.  31.  1.)  AF 
parallel  to  CD  or  BE  ;  then  AE=AD 
+  CE. 

But  AE  =  AB.BE  =  AB.BC,  be- 
cause BE=BC.  So  also  AD=:AC. 
CD=:AC.CB;  and  CE=BC2;  there- 
fore AB.BC=AC.CB+BC2. 

SCHOLIUM. 

In  this  proposition  let  AB  be  denoted  by  a,  and  the  segments  AC  and 
CB,  by  h  and  c  ;  then  a=&-fc  :  therefore,  multiplying  both  members  of 
this  equality  by  c,  we  shall  have  ac=6c+c^. 


PROP.  IV.    THEOR. 

If  a  straight  line  he  divided  into  any  two  parts,  the  square  of  the  whole  line  is 
equal  to  the  squares  of  the  two  parts,  together  with  twice  the  rectangle  con- 
tained  by  the  parts. 

Let  the  straight  line  AB  be  divided  into  any  two  parts  in  C  ;  the  square 
of  AB  is  equal  to  the  squares  of  AC,  CB,  and  to  twice  the  rectangle  con- 
tained by  AC,  CB,  that  is,  AB2=AC2+CB2-f  ^AC.CB. 

Upon  AB  describe  (Prop.  46.  1.)  the  square  ADEB,  and  join  BD,  and 
through  C  draw  (Prop.  31.  1.)  CGF  parallel  to  AD  or  BE,  and  through  G 
draw  HK  parallel  to  AB  or  DE.  And  because  CF  is  parallel  to  AD,  and 
BD  falls  upon  them,  the  exterior  angle  BGC 
is  equal  (29.  1.)  to  the  interior  and  opposite 
angle  ADB  ;  but  ADB  is  equal  (5.  1.)  to  the 
angle  ABD,  because  BA  is  equal  to  AD,  be- 
ing sides  of  a  square  ;  wherefore  the  angle 
CGB  is  equal  to  the  angle  GBC  ;  and  there- 
fore the  side  BC  is  equal  (6.  1.)  to  the  side 
CG  ;  but  CB  is  equal  (34.  1.)  also  to  GK  and 
CG  to  BK ;  wherefore  the  figure  CGKB  is 
equilateral.  It  is  likewise  rectangular ;  for 
the  angle  CBK  being  a  right  angle,  the  other 
angles  of  the  parallelogram  CGKB  are  also  right  angles  (Cor.  46.  1.) 
Wherefore  CGKB  is  a  square,  and  it  is  upon  the  side  CB.     For  the  same 


A             C        ] 

B 

H 

Ct 

IT 

/ 

Ak 

I 

)                          ] 

?            ] 

E 

OF  GEOMETRY.     BOOK  II. 


51 


reason  HF  also  is  a  square,  and  it  is  upon  the  side  HG,  which  is  equal  to 
AC  :  therefore  HF,  CK  are  the  squares  of  AC,  CB.  And  because  the 
complement  AG  is  equal  (43.  1.)  to  the  complement  GE  ;  and  because 
AG=AC.CG=AC.CB,  therefore  also  GE=AC.CB,  and  AG+GE=: 
2AC.CB,  Now,  HF=AC2  and  CK=CB2;  therefore,  HF+CK+AG 
-f  GE=AC2+CB2+2AC.CB. 

But  HF+CK+AG4-GE=the  figure  AE,  or  AB2;  therefore  AB2= 
AC2-fCB2+2AC.CB. 

CoR.  From  the  demonstration,  it  is  manifest  that  the  parallelograms 
about  the  diameter  of  a  square  are  likewise  squares. 

SCHOLIUM. 

This  property  is  derived  from  the  square  of  a  binomial.  For,  let  the  two 
parts  into  which  this  line  is  divided  be  denoted  by  a  and  b ;  then,  (a-\-b)^ 
=a2+2<z6-fR 

PROP.  V.    THEOR. 


If  a  straight  linele  dividedinto  two  equal  parts,  and  also  into  two  unequal  parts  ; 
the  rectangle  contained  hy  the  unequal  parts,  together  with  the  square  of  the 
line  between  the  points  of  section^  is  equal  to  the  square  of  half  the  line. 

Let  the  straight  line  AB  be  divided  into  two  equal  parts  in  the  point  C,^ 
and  into  two  unequal  parts  in  the  point  D  ;  the  rectangle  AD.DB,  together^ 
with  the  square  of  CD,  is  equal  to  the  square  of  CB,  or  AD.DB-fCD^ 
CB2. 

Upon  CB  describe  (Prop.  46. 1.)  the  square  CEFB,  join  BE,  and  throu 
D  draw  (Prop.  31. 1.)  DHG  parallel  to  CE  or  BF  ;  and  through  H  dra 
KLM  parallel  to  CB  or  EF  ;  and  ^ 


also  through  A  draw  AK  parallel  to 
CLor  BM  :  And  because  CH=HF, 
if  DM  be  added  to  both,  CM=DF. 
But  AL=(36. 1.)  CM,  therefore  AL 
=DF,  and  adding  CH  to  both,  AH 
=gnomon  CMG.  But  AH  =  AD. 
DH=AD.DB,  because  DH  =  DB 
(Cor.  4.  2.) ;  therefore  gnomon  CMG 


k: 


T)      B 


:ei 


II 


M 


Cr     F 


=AD.DB.     To  each  add  LG=CD2,  then,  gnomon  CMG4-LG=AD.DB 
+  CD2.     But  CMG+LG=BC2;  therefore  AD.DB  +  CD2=BC2. 

"  Cor.  From  this  proposition  it  is  manifest,  that  the  difference  of  the 
*'  squares  of  two  unequal  lines,  AC,  CD,  is  equal  to  the  rectangle  contain- 
**  ed  by  their  sum  and  difference,  or  that  AC2_CD2=(AC  +  CD)  (AC — 
**  CD)." 

SCHOLIUM. 

In  this  proposition,  let  AC  be  denoted  by  «,  and  CD  by  h  ;  then,  AD= 
a-\-b,  and  DB=a — b\  therefore,  by  Algebra,  [a-{-b)x(a — b)=a'^ — 5^  . 
that  is,  the  product  of  the  sum  and  difference  of  two  quantities,  is  equivalent 
to  the  difference  of  their  squares 


52 


ELEMENTS 


PROP.  VI  THEOR. 

If  a  straight  line  he  bisected,  and  produced  to  any  point ;  the  rectangle  contained 
hy  the  whole  line  thus  produced,  and  the  part  of  it  produced,  together  with  the 
square  of  half  the  line  bisected,  is  equal  to  the  square  of  the  straight  line  which 
is  made  up  of  the  half  and  the  part  produced. 

Let  the  straight  line  AB  be  bisected  in  C,  and  produced  to  the  point  D  ; 
the  rectangle  AD.DB  together  with  the  square  of  CB,  is  equal  to  the 
square  of  CD. 

Upon  CD  describe  (Prop.  46.1.)  the  square  CEFD,  join  DE,  and 
through  B  draw  (Prop.  31.1.)  BHG  parallel  to  CE  or  DF,  and  through  H 
draw  KLM  parallel  to  AD  or  EF,  and  also  through  A  draw  AK  parallel 
to  CL  or  DM.     And  because  AC  is 
equal  to  CB,  the  rectangle  AL  is 
equal  (36.1.)  to  CH ;  but  CH  is 
equal  (43. 1. )  to  HF  ;  therefore  also 
AL  is  equal  to  HF :  To  each  of  these 
add  CM ;  therefore  the  whole  AM  is 
equal  to  the  gnomon  CMC     Now 
AM=AD.DM  =  AD.DB,  because 
DM=DB.  Therefore  gnomon  CMG 
=AD.DB,  and  CMG+LG=AD. 
DB+CB2.     But  CMG+LG=CF 
=CD2,  therefore  AD.DB+CB2=CD2. 

SCHOLIUM. 
This  property  is  evinced  algebraically ;  thus,  let  AB  be  denoted  by  2a, 
and  BD  by  i> ;  then,  AD=2a+6,  and  CD=a+*.     Now  by  multiplication, 

5f2a+i)=2a54-52  ;  therefore, 
by  adding  a^  to  each  member  of  the  equality,  we  shall  have, 

i(2c+6)+a2=aH2ai+^^  ; 
.•.i'(2c-|-&)+c2=(a+i)2. 

PROP.  VII.    THEOR. 

Jfa  straight  line  he  divided  into  two  parts,  the  squares  of  the  whole  line,  and 
of  one  ^  the  parts,  are  equal  to  twice  the  rectangle  contained  by  the  whole  and 
that  part,  together  with  the  square  of  the  other  part. 

C      B 

K 


G    P 


L  et  the  straight  line  AB  be  divided  into  any 
two  parts  in  the  point  C  ;  the  squares  of  AB, 
BC,  are  equal  to  twice  the  rectangle  AB.BC, 
together  with  the  square  of  AC,  or  AB^+BC^ 
=2AB.BC4-AC2. 

Upon  AB  describe  (Prop.  46. 1.)  the  square 
ADEB,  and  construct  the  figure  as  in  the  pre- 
ceding propositions  :  Because  AG=GE,  AG 
+  CK  =  GE-f-CK,  that  is,  AK  =  CE,  and 
therefore  AK4-CE=2AK.  But  AK+CE 
=gnomon   AKF+CK;  and   therefore  AKF 


H 


D 


]?       E 


OF  GEOMETRY.    BOOK  II. 


53 


-|-CK=2AK  =  2AB.BK  =  2AB.BC,  because  BK  =  (Cor.  4.  2.)  BC. 
Since  then,  AKF-hCK=2AB.BC,  AKF+CK+HF=2AB.BC+HF ; 
and  because  AKF+HF=:AE=AB2,  AB24.CK=2AB.BC+HF,  that 
is,  (since  CK=CB2,  and  HFr=AC2,)  AB2+CB2=^2AB.BC4-AC2. 

"  Cor.  Hence,  the  sum  of  the  squares  of  any  two  lines  is  equal  to  1 
"  twice  the  rectangle  contained  by  the  lines  together  with  the  square  of  J 
"  the  difference  of  the  lines."  / 

SCHOLIUM. 

In  this  proposition,  let  AB  be  denoted  by  a,  and  the  segments  AC  and 
CB  by  J  and  c ; 

then  a2=52_^25c+c2 ; 
adding  c^  to  each  member  of  this  equality,  we  shall  have, 
a2^c2=:62+25c+l?c2; 
/.  a24-c2=i2_{.2c(6+c), 
or  a2+c2=:2ac+62. 

Cor.  From  this  proposition  it  is  evident,  that  the  square  described  on 
the  difference  of  two  lines  is  equivalent  to  the  sum  of  the  squares  described  on 
the  lines  respectively^  minus  twice  the  rectangle  contained  by  the  lines.  For 
a — c=ib  ;  therefore,  by  involution,  a^ — 2ac-fc2=i2,  This  may  be  also 
derived  from  the  above  algebraical  equality,  by  transposition. 


PROP.  VIII.    THEOR. 


If  a  straight  line  be  divided  into  any  two  parts,  four  times  the  rectangle  con- 
tainedby  the  whole  line,  and  one  of  the  parts,  together  with  the  square  of 
the  other  part,  is  equal  to  the  square  of  the  straight  line  which  is  made  up 
of  the  whole  and  the  first-mentioned  part. 

Let  the  straight  line  AB  be  divided  into  any  two  parts  in  the  point  C ; 
four  times  the  rectangle  AB.BC,  together  with  the  square  of  AC,  is  equal 
to  the  square  of  the  straight  line  made  up  of  AB  and  BC  together. 

Produce  AB  to  D,  so  that  BD  be  equal  to  CB,  and  upon  AD  describe 
the  square  AEFD ;  and  construct  two  figures  such  as  in  the  preceding. 
Because  GK  is  equal  (34.  1.)  to  CB,  and  CB  to  BD,  and  BD  to  KN,  GK 
is  equal  to  KN.  For  the  same  reason,  PR 
is  equal  to  RO  ;  and  because  CB  is  equal 
to  BD,  and  GK  to  KN,  the  rectangles  CK 
and  BN  are  equal,  as  also  the  rectangles 
GR  and  RN :  But  CK  is  equal  (43.  1.) 
to  RN,  because  they  are  the  complements 
of  the  parallelogram  CO  :  therefore  also 
BN  is  equal  to  GR  ;  and  the  four  rect- 
angles BN,  CK,  GR,  RN  are  there- 
fore equal  to  one  another,  and  so  CK-f- 
BN  +  GR  +  RN  =  4CK.  Again,  be- 
cause CB  is  equal  to  BD,  and  BD  equal       Jii  H     li     JF 


C    B     B 


M 
X 

E 


G 

K 

/ 

P 

/ 

R 

/ 

N 
O 


54 


ELEMENTS 


(Cor.  4.  2.)  to  BK,  that  is,  to  CG ;  and  CB  equal  to  GK,  that  is,  to  GP ; 
therefore  CG  is  equal  to  GP  ;  and  because  CG  is  equal  to  GP,  and  PR  to 
RO,  the  rectangle  xiG  is  equal  to  MP,  and  PL  to  RF :  but  MP  is  equal 
(43.  1.)  to  PL,  because  they  are  the  complements  of  the  parallelogram 
ML ;  wherefore  AG  is  equal  also  to  RF.  Therefore  the  four  rectangles 
AG,  MP,  PL,  RF,are  equal  to  one  another,  and  so  AG+MP+PL+RF 
=4AG.  And  it  was  demonstrated,  that  CK+BN+GR+RN=4CK  ; 
wherefore,  adding  equals  to  equals,  the  whole  gnomon  A0H=4AK. 
Now  AK=AB.BK=AB.BC,  and  4AK=4AB.BC  ;  therefore,  gnomon 
A0H=4AB.BC ;  and  adding  XH,  or  (Cor.  4.  2.)  AC2,  to  both,  gnomon 
AOH4-XH=4AB.BC+AC2.  But  A0H+XH=AF  =  AD2;  therefore 
AD2=4AB.BC+AC2. 

"  Cor.  I.  Hence,  because  AD  is  the  sum,  and  AC  the  difference  of 
"  the  lines  AB  and  BC,  four  times  the  rectangle  contained  by  any  two 
"  lines,  together  with  the  square  of  their  difference,  is  equal  to  the  square 
"  of  the  sum  of  the  lines." 

"  Cor.  2.  From  the  demonstration  it  is  manifest,  that  since  the  square 
"  of  CD  is  quadruple  of  the  square  of  CB,  the  square  of  any  line  is  qua* 
"  druple  of  the  square  of  half  that  line." 

SCHOLIUM. 

In  this  proposition,  let  the  line  AB  be  denoted  by  a,  and  the  parts  AC 
and  CB  by  c  and  b ;  then  AD=c+26.  Now,  since  a—b-{-Cf  multiplying 
both  members  by  45,  we  shall  have 

4ab=4b'2-{-4bc; 
and  adding  c^  to  each  member  of  this  equality,  we  shall  have, 
4a54-c2=c2+4^>c+462, 
or  4a6+c2=(c  4-26)2. 

PROP.  IX.    THEOR. 


If  a  straight  line  be  divided  into  two  equal,  and  also  into  two  unequal  parts , 
the  squares  of  the  two  unequal  parts  are  together  double  of  the  square  of  half 
the  line,  and  of  the  square  of  the  line  between  the  points  of  section. 

Let  the  straight  line  AB  be  divided  at  the  point  C  into  two  equal,  and 
at  D  into  two  unequal  parts  ;  The  squares  of  AD,  DB  are  together  double 
of  the  squares  AC,  CD. 

From  the  point  C  draw  (Prop.  11.1.)  CE  at  right  angles  to  AB,  and 
make  it  equal  to  AC  or  CB,  and  join  EA,  EB  ;  through  D  draw  (Prop.  31. 
1.)  DF  parallel  to  CE,  and  through  F  draw  FG  parallel  to  AB  ;  and  join 
AF.  Then,  because  AC  is  equal  to  CE, 
the  angle  EAC  is  equal  (5.  1.)  to  the 
angle  AEC ;  and  because  the  angle  ACE 
is  a  right  angle,  the  two  others  AEC, 
EAC  together  make  one  right  angle  (Cor. 
4.  32. 1 .) ;  and  they  are  equal  to  one  ano- 
ther ;  each  of  them  therefore  is  half  of  a 
right  angle.     For  the  same  reason  each 


OF  GEOMETRY.     BOOK  11. 


55 


of  the  angles  CEB,  EBC  is  half  a  right  angle ;  and  therefore  the  whole 
AEB  is  a  right  angle  ;  And  because  the  ^ngle  GEF  is  half  a  right  angle, 
and  EGF  a  right  angle,  for  it  is  equal  (29.  1.)  to  the  interior  and  opposite 
angle  ECB,  the  remaining  angle  EFG  is  half  a  right  angle  ;  therefore  the 
angle  GEF  is  equal  to  the  angle  EFG,  and  the  side  EG  equal  (6.  1.)  to  the 
side  GF  ;  Again,  because  the  angle  at  B  is  half  a  right  angle,  and  FDB  a 
right  angle,  for  it  is  equal  (29.  1.)  to  the  interior  and  opposite  angle  ECB, 
the  remaining  angle  BFD  is  half  a  right  angle  ;  therefore  the  angle  at  B  is 
equal  to  the  angle  BFD,  and  the  side  DF  to  (6.  1 .)  the  side  DB.  Now,  be- 
cause AC=CE,  AC2=CE2,  and  AC2+CE2=2AC2.  But  (47. 1.)  AE'^rr: 
AC24-CE2 ;  therefore  AE2=2AC2.  Again, because EG=GF,  EG2=GF2, 
and  EG2+GF2=2GF2.  But  EF2=:EG2+GF2 ;  therefore,  EF2=2GF2 
=2CD2,  because  (34. 1.)  CD  =  GF.  And  it  was  shown  that  AE2=2AG2 ; 
therefore  AE2+EF2==2AC2+2CD2.  But  (47.  1.)  AF2=AE24-EF2, 
and  AD2+DF2=: AF2, or  AD2+DB2=AF2 ;  therefore,  also,  AD2-|-DB2= 
2AC2+2CD2. 

SCHOLIUM. 
This  property  is  evident  from  the  algebraical  expression, 
(a+i)2+(a— 5)2=2a24-2i2 ; 
where  a  denotes  AC,  and  h  denotes  CD  ;  hence,  a-\-h  =AD,  a — i=DB. 

PROP.  X.     THEOR. 


If  a  straight  line  he  bisected,  and  produced  to  any  point,  the  square  of  the  whole 
line  thus  produced,  and  the  square  of  the  part  of  it  produced,  are  together 
double  of  the  square  of  half  the  line  bisected,  and  of  the  square  of  the  line 
made  up  of  the  half  and  the  part  produced. 

Let  the  straight  line  AB  be  bisected  in  C,  and  produced  to  the  point  D ; 
the  squares  of  AD,  DB  are  double  of  the  squares  of  AC,  CD. 

From  the  point  C  draw  (Prop.  11.1.)  CE  at  right  angles  to  AB,  and  make 
it  equal  to  AC  or  CB  ;  join  AE,  EB ;  through  E  draw  (Prop.  31.  1.)  EF 
parallel  to  AB,  and  through  D  draw  DF  parallel  to  CE.  And  because 
the  straight  line  EF  meets  the  parallels  EC,  FD,  the  angles  GEF,  EFD 
are  equal  (29. 1.)  to  two  right  angles  ;  and  therefore  the  angles  BEF,  EFD 
are  less  than  two  right  angles ;  But  straight  lines,  which  with  another 
straight  line  make  the  interior  angles  upon  the  same  side  less  than  two 
right  angles,  do  meet  (29.  1.),  if  produced  far  enough ;  therefore  EB,  FD 
will  meet,  if  produced,  towards  B,  D  :  let  them  meet  in  G,  and  join  AG. 
Then  because  AC  is  equal  to  CE, 
the  angle  CEA  is  equal  (5.1.)  to 
the  angle  EAC ;  and  the  angle 
ACE  is  a  right  angle  ;  therefore 
each  of  the  angles  CEA,  EAC  is 
half  a  right  angle  (Cor.  4.  32.  1.); 
For  the  same  reason,  each  of  the 
angles  CEB,  EBC  is  half  a  right 
angle;  therefore  AEB  is  a  right  an- 
gle ;  And   because  EBC  is  half  a 


56 


ELEMENTS 


right  angle,  DBG  is  also  (15.  1.)  half  a  right  angle,  for  they  are  vertically 
opposite  :  but  BDG  is  a  right  angle,  because  it  is  equal  (29.  1.)  to  the  al- 
ternate angle  DCE  ;  therefore  "the  remaining  angle  DGB  is  half  a  right 
angle,  and  is  therefore  equal  to  the  angle  DBG;  wherefore  also  the  side 
DB  is  equal  (6.  1.)  to  the  side  DG.  Again,  because  EGF  is  half  a  right 
angle,  and  the  angle  at  F  aright  angle,  being  equal  (34.  1.)  to  the 
opposite  angle  ECD,  the  remaining  angle  FEG  is  half  a  right  angle, 
and  equal  to  the  angle  EGF;  wherefore  also  the  side  GF  is  equal 
(6.  1.)  to  the  side  FE.  And  because  ECzzzCA,  EG2  +  CA^  ^  2CA2. 
Now  AE2r=  (47.  1.)  AC2 -I- CE2;  therefore,  AE2=  2AC2.  Again,  be- 
cause EF=FG,  EF2=FG2,  andEF2+FG2=:2EF2.  ButEG2:rr  (47.  1.) 
EF24.FG2;  therefore  EG2=:2EF2;  and  since  EF=CD,  EG2=:2CD2. 
And  it  was  demonstrated,  that  AE2=2AC2  ;  therefore,  AE2+EG2=2AC2 
+2CD2.  Now,  AG2=AE2+EG2,  wherefore  AG2=2AC2+2CD2.  But 
AG2(47.  l.)  =  AD24-DG2=AD2-f-DB2,  because  DG=DB :  Therefore, 
AD2+DB2=:2AC2-f2CD2. 

SCHOLIUM. 

Let  AC  be  denoted  by  a,  and  BD,  the  part  produced,  by  b  ;  then  KDzrz 
2a-\rh,  and  CV>=a-{-h. 

Now,  (2aH-6)2-f  62^4a2-|.4a6+26^;  but  4a2  +  4a5+2P=2a2+2  (a-f 
5)2;  hence,  (2a4-6)2-|-62—2a2_|-2(a-{- 5)2,  and  the  proposition  is  evident 
from  this  algebraical  equality. 

PROP.  XL     PROB. 

To  divide  a  given  straight  line  into  two  parts,  so  that  the  rectangle  contained 
hy  the  whole,  and  one  of  the  parts,  may  he  equal  to  the  square  of  the  other 
parti 


Let  AB  be  the  given  straight  line ;  .it  is  required  to  divide  it  into  two 


parts,  so  that  the  rectangle  contained  by 
the  whole,  and  one  of  the  parts,  shall  be 
equal  to  the  square  of  the  other  part. 

Upon  AB  describe  (46.  1.)  the  square 
ABDC ;  bisect  (10.  1.)  AC  in  E,  and  join 
BE  ;  produce  CA  to  F,  and  make  (3.  1.) 
EF  equal  to  EB,  and  upon  AF  describe 
(46.  1.)  the  square  FGHA,  AB  is  divided 
in  H,  so  that  the  rectangle  AB,  BH  is  equal 
to  the  square  of  AH. 

Produce  GH  to  K  :  Because  the  straight 
line  AC  is  bisected  in  E,  and  produced  to 
the  point  F,  the  rectangle  CF.FA,  to- 
gether Avith  the  square  of  AE,  is  equal 
(6.  2.)  to  the  square  of  EF:  But  EF  is 
equal  to  EB  ;  therefore  the  rectangle  CF. 
FA,  together  with  the  square  of  AE,  is 


OF  GEOMETRY.     BOOK  II. 


57 


equal  to  the  square  of  EB  ;  And  the  squares  of  BA,  AE  are  equal 
(47.  l.)to  the  square  of  EB,  because  the  angle  EAB  is  a  right  angle; 
therefore  the  rectangle  CF.FA,  together  with  the  square  of  AE,  is  equal 
to  the  squares  of  BA,  AE  :  take  away  the  square  of  AE,  which  is  com- 
mon to  both,  therefore  the  remaining  rectangle  CF.FA  is  equal  to  the 
square  of  AB.  Now  the  figure  FK  is  the  rectangle  CF.FA,  for  AF  is 
equal  to  FG  ;  and  AD  is  the  square  of  AB  ;  therefore  FK  is  equal  to  AD  ; 
take  away  the  common  part  AK,  and  the  remainder  FH  is  equal  to  the 
remainder  HD.  But  HD  is  the  rectangle  AB.BH  for  AB  is  equal  to 
BD ;  and  FH  is  the  square  of  AH  ;  therefore  the  rectangle  AB.BH  is 
equal  to  the  square  of  AH  :  Wherefore  the  straight  line  AB  is  divided  in 
H,  so  that  the  rectangle  AB.BH  is  equal  to  the  square  of  AH. 


PROP.  XII.    THEOR. 

In  obtuse  angled  triangles,  if  a  perpendicular  he  drawn  from  any  of  the  acute 
angles  to  the  opposite  side  produced,  the  square  of  the  side  subtending  the 
obtuse  angle  is  greater  than  the  squares  of  the  sides  containing  the  obtuse 
angle,  by  twice  the  rectangle  contained  by  the  side  uponwhich, when  produced, 
the  perpendicular  falls,  and  the  straight  line  intercepted  betioeen  the  perpen- 
dicular and  the  obtuse  angle. 

Let  ABC  be  an  obtuse  angled  triangle,  having  the  obtuse  angle  ACB, 
and  from  the  point  A  let  AD  be  drawn  (12.  1.)  perpendicular  to  BC  pro- 
duced :  The  square  of  AB  is  greater  than  the  squares  of  AC,  CB,  by  twice 
the  rectangle  BC.CD. 

Because  the  straight  line  BD  is  divided  .A. 

into  two  parts  in  the  point  C,  BD2  =  (4.  2.) 
BC2+CD2+2BC.CD;  add  AD^  to  both: 
Then  BD2+AD2  =  BC2+  CD2+  AD2-f- 
2BC.CD.  But  AB2=BD2-f  AD2(47.  1.), 
and  AC2=  CD2+ AD2  (47.  1.);  therefore, 
AB2=BC2+AC2+2BC.CD;  that  is,  AB2 
is  greater  than  BC2+AC2  by  2BC.CD. 


PROP.  XIII.     THEOR. 


In  every  triangle  the  square  of  the  side  subtending  any  of  the  acute  angles,  is 
less  than  the  squares  of  the  sides  containing  that  angle,  by  twice  the  rectan- 
gle contained  by  either  of  these  sides,  and  the  straight  line  intercepted  be- 
tween the  perpendicular,  let  fall  upon  it  from  the  opposite  angle,  and  the  acute 
angle. 

Let  ABC  be  any  triangle,  and  the  angle  at  B  one  of  its  acute  angles,  and 
upon  BC,  one  of  the  sides  containing  it,  let  fall  the  perpendicular  (12.  1.) 
AD  from  the  opposite  angle  :  The  square  of  AC,  opposite  to  the  angle  B, 
is  less  than  the  squares  of  CB,  BA  by  twice  the  rectangle  CB.BD. 

8 


58 


ELEMENTS 


First, let  AD  fall  within  the  triangle  ABC ; 
and  because  the  straight  line  CB  is  divided 
into  two  parts  in  the  point  D  (7.  2.),  BC^-f 
BD2=.2BC.BD  +  CD2.  Addtoeach  AD2; 
thenBC2+BD2-f  AD2=2BC.BD  +  CD2-1- 
AD2.  But  BD2-|-AD2.3:AB2,  and  CD2+ 
DA2=:  AC2  (47. 1.) ;  therefore  BC2+AB2  — 
2BC.BD  +  AC2 .  that  is,  AC'^  is  less  than 
BC2+AB2by2BC.BD. 


B 


D  C 

Secondly,  let  AD  fall  without  the  triangle  ABC  :*  Then  because  the 
angle  at  D  is  a  right  angle,  the  angle  ACB  is  greater  (16.  I.)  than  a  right 
angle,  and  AB^^  (12.  2.)  AC^+BC24-2BC.CD.  Add  BC2  to  each; 
then  AB2+BC2—AC2-I-2BC2+2BC.CD.  But  because  BD  is  divided 
into  two  parts  in  C,  BC2+BC.CD=(3.  2.)  BC.BD,  and  2BC2-f  2BC.CD 
r=2BC.BD:  therefore  AB2+ BC2=:AC2+ 2BC.BD ;  and  AC2  isless 
than  AB2+BC2,  by  2BD.BC. 


Lastly,  let  the  side  AC  be  perpendicular 
to  BC  ;  then  is  BC  the  straight  line  between 
the  perpendicular  and  the  acute  angle  at  B  ; 
and  it  is  manifest  that  (47.  1.)  AB2+BC2= 
AC2+2BC2=AC2-1-2BC.BC. 


PROP.  XIV.    PROB. 


To  describe  a  square  that  shall  he  equal  to  a  given  rectilineal  figure. 

Let  A  be  the  given  rectilineal  figure  ;  it  is  required  to  describe  a  square 
that  shall  be  equal  to  A. 

Describe  (45.  1.)  the  rectangular  parallelogram  BCDE  equal  to  the 
rectilineal  figure  A.  If  then  the  sides  of  it,  BE,  ED  are  equal  to  one  an- 
other, it  is  a  square,  and  v/hat  was  required  is  done ;  but  if  they  are  not 
equal,  produce  one  of  them,  BE  to  F,  and  make  EF  equal  to  ED,  and  bi- 
sect ]BF  in  G ;  and  from  the  centre  G,  at  the  distance  GB,  or  GF,  de- 
scribe the  semicircle  BHF,  and  produce  DE  to  H,  and  join  GH.  There- 
fore, because  the  straight  line  BF  is  divided  into  two  equal  parts  in  the 
point  G,  and  into  two  unequal  in  the  point  E,  the  rectangle  BE.EF,  to- 
gether with  the  square  of  EG,  is  equal  (5.  2.)  to  the  square  of  GF  : 
but  GF  is  equal  to  GH  ;  therefore  the  rectangle  BE,  EF,  together 
with  the  square  of  EG,  is  equal  to  the  square  of  GH  :  But  the  squares  of 


*  See  figure  of  the  last  Proposition. 


OF  GEOMETRY.    BOOK  II. 


59 


HE  and  EG  are  equal  (47. 
1.)  to  the  square  of  GH  : 
Therefore  also  the  rectangle 
BE.EF,  together  with  the 
square  of  EG,  is  equal  to 
the  squares  of  HE  and  EG. 
Take  away  the  square  of 
EG,  which  is  common  to 
both,  and  the  remaining 
rectangle  BE.EF  is  equal 
to  the  square  of  EH  :  But 
BD  is  the  rectangle  con- 
tained by  BE  and  EF,  because  EF  is  equal  to  ED  ;  therefore  BD  is  equal 
to  the  square  of  EH  ;  and  BD  is  also  equal  to  the  rectilineal  figure  A  ; 
therefore  the  rectilineal  figure  A  is  equal  to  the  square  of  EH  :  Where- 
fore a  square  has  been  made  equal  to  the  given  rectilineal  figure  A,  viz. 
the  square  described  upon  EH. 

PROP.  A.    THEOR. 

If  one  side  of  a  triangle  he  bisected,  the  sum  of  the  squares  of  the  other  two 
sides  is  double  of  the  square  of  half  the  side  bisected,  and  of  the  square 
of  the  line  drawn  from  the  point  of  bisection  to  the  opposite  angle  of  the 
triangle. 

Let  ABC  be  a  triangle,  of  which  the  side  BC  is  bisected  in  D,  and  DA 
drawn  to  the  opposite  angle  ;  the  squares  of  BA  and  AC  are  together 
double  of  the  squares  of  BD  and  DA. 

From  A  draw  AE  perpendicular  to  BC,  and  because  BEA  is  a  right  an- 
gle, AB2=(47. 1.)  BE2+AE2  and  AC2== 

CE24.  AE2 ;  wherefore  AB^+AC^^BE^  A 

+  CE2+2AE2.  But  because  the  line 
BC  is  cut  equally  in  D,  and  unequally 
in  E,  BE2  +  CE2  =  (9.  2.)  2BD2  + 
2DE2 ;  therefore  AB^  +  AC2=2BD2  -f 
2DE2.2AE2. 

Now  DE24-AE2=(47.  1.)  AD2,  and 
2DE2-f2AE2=:2AD2;  wherefore  AB2-|- 
AC2=2BD2-f2AD2. 


PROP.  B.    THEOR. 

The  sum  of  the  squares  of  the  diameters  of  any  parallelogram  is  equal  to 
the  sum  of  the  squares  of  the  sides  of  the  parallelogram. 

Let  ABCD  be  a  parallelogram,  of  which  the  diameters  are  AC  and  BD  ; 
the  sum  of  the  squares  of  AC  and  BD  is  equal  to  the  sum  of  the  squares 
of  AB,  BC,  CD,  DA. 

Let  AC  and  BD  intersect  one  another  in  E  :  and  because  the  vertical 
angles  AED,  CEB  are  equal  (15.  L),  and  also  the  alternate  angles  EAD, 


60  ELEMENTS,  &c. 

ECB  (29.  1.),  the  triangles  ADE,  CEB  have  two  angles  in  the  one  equal 

to  two  angles  in  the  other,  each  to  each ;  but  the  sides  AD  and  BC,  which 

are  opposite  to  equal  angles  in 

these  triangles,  are  also  equal 

(34.    1.);  therefore    the   other 

sides  which  are  opposite  to  the 

equal  angles  are  also  equal  (26. 

1.),  viz.  AE  to  EC,  and  ED  to 

EB. 

Since,  therefore,  BD  is  bi- 
sected in  E,  AB2+AD2=(A. 
2.)  2BE24-2AE2;  and  for  the 
same  reason,  CD^  +  BC^  = 
2BE24-2EC2=2BE2-f2AE2,  because  EC=AE.  Therefore  AB2+AD2 
H-DC2+BC2=4BE2+4AE2.  But  4BE2=BD2,  and  4AE2=AC2  (2. 
Cor.  8.  2.)  because  B  D  and  AC  are  both  bisected  in  E  ;  therefore  AB24- 
AD2-fCD2+BC2=BD2+AC2. 

Cor.  From  this  demonstration,  it  is  manifest  that  the  diameters  of  every 
parallelogram  bisect  one  another. 

SCHOLIUM. 

In  the  case  of  the  rhombus,  the  sides  AB,  BC,  being  equal,  the  triangles 
BEC,  DEC,  have  all  the  sides  of  the  one  equal  to  the  corresponding  sides 
of  the  other,  and  are  therefore  equal :  whence  it  follows  that  the  angles 
BEC,  DEC,  are  equal ;  and,  therefore,  that  the  two  diagonals  of  a  rhom- 
bus cut  each  other  at  right  angles. 


ELEMENTS 

OF 

GEOMETRY. 


BOOK  III. 

DEFINITIONS. 


A.  The  radius  of  a  circle  is  the  straight  line  drawn  from  the  centre  to  the 
circumference. 

1.  A  straight  line  is  said  to  touch 
a  circle,  when  it  meets  the  cir- 
cle, and  being   produced  does 
not  cut  it. 
And   that   line  which    has  but 

one  point  in  common  with 
the  circumference,  is  called  a 
tangent^  and  the  point  in  com- 
mon, the  point  of  contact, 

2.  Circles  are  said  to  touch  one 
another,  which  meet,  but  do  not 
cut  one  another. 

3.  Straight  lines  are  said  to  be  equally  dis- 
tant from  the  centre  of  a  circle,  when  the 
perpendiculars  drawn  to  them  from  the  centre 
are  equal. 

4.  And  the  straight  line  on  which  the  greater 
perpendicular  falls,  is  said  to  be  farther  from 
the  centre. 

B.  Any  portion  of  the  circumference  is  called  an  arc. 
The  chord  or  subtense  of  an  arc  is  the  straight  line  which  joins  its  two  ex- 
tremities. 

C.  A  straight  line  is  said  to  be  inscriledin  a  circle,  when  the  extremities  of 
it  are  in  the  circumference  of  the  circle.  And  any  straight  line  which 
meets  the  circle  in  two  points,  is  called  a  secant. 

5-  A  segment  of  a  circle  is  the  figure  con- 
tained by  a  straight  line,  and  the  arc  which 
it  cuts  off. 


62 


ELEMENTS 


6.  An  angle  in  a  segment  is  the  angle  contained 
by  two  straight  lines  drawn  from  any  point  in 
the  circumference  of  the  segment,  to  the  extre- 
mities of  the  straight  line  which  is  the  base  of 
the  segment. 

An  inscribed  triangle,  is  one  which  has  its  three 
angular  points  in  the  circumference. 

And,  generally,  an  inscribed  figure  is  one,  of 
which  all  the  angles  are  in  the  circumference. 
The  circle  is  said  to  circumscribe  such  a  figure. 

7.  And  an  angle  is  said  to  insist  or  stand  upon 
the  arc  intercepted  between  the  straight  lines 
which  contain  the  angle. 

This  is  usually  called  an  angle  at  the  centre.  The 
angles  at  the  circumference  and  centre,  are 
both  said  to  be  subtended  by  the  chords  or 
arcs  which  their  sides  include. 

8.  The  sector  of  a  circle  is  the  figure  contained 
by  two  straight  lines  drawn  from  the  centre,  and 
the  arc  of  the  circumference  between  them. 


9.  Similar  segments  of  a  circle, 
are  those  in  which  the  angles  are 
equal,  or  which  contain  equal  an- 
gles. 


PROP.  I.     PROB. 
To  find  the  centre  of  a  given  circle. 

Let  ABC  be  the  given  circle  ;  it  is  required  to  find  its  centre. 

Draw  within  it  any  straight  line  AB,  and  bisect  (10.  1.)  it  in  D  ; 
from  the  point  D  draw  (11.  1.)  DC  at  right  angles  to  AB,  and  produce  it 
to  E,  and  bisect  CE  in  F  :  the  point  F  is  the  centre  of  the  circle  ABC. 

For,  if  it  be  not,  let,  if  possible,  G  be  the  centre,  and  join  GA,  GD,  GB  ; 
Then,  because  DA  is  equal  to  DB,  and  DG  common  to  the  two  triangles 
ADG,  BDG,  the  two  sides  AD,  DG  are  equal  to 
the  two  BD,  DG,  each  to  each  ;  and  the  base 
GA  is  equal  to  the  base  GB,  because  they  are 
radii  of  the  same  circle  :  therefore  the  angle 
ADG  is  equal  (8.  1.)  to  the  angle  GDB :  But 
when  a  straight  line  standing  upon  another 
straight  line  makes  the  adjacent  angles  equal  to 
one  another,  each  of  the  angles  is  a  right  angle 
(7.  def.  1 .)  Therefore  the  angle  GDB  is  a  right 
angle :  But  FDB  is  likewise  a  right  angle : 
wherefore  the  angle  FDB  is  equal  to  the  angle 
GDB,  the  greater  to  the  less,  which  is  impos- 


OF  GEOMETRY.     BOOK  III. 


63 


sible :  Therefore  G  is  not  the  centre  of  the  circle  ABC  :  In  the  same 
manner,  it  can  be  shown  that  no  other  point  but  F  is  the  centre  :  that  is, 
F  is  the  centre  of  the  circle  ABC. 

Cor.  From  this  it  is  manifest  that  if  in  a  circle  a  straight  line  bisect 
another  at  right  angles,  the  centre  of  the  circle  is  in  the  line  which  bisects 
the  other. 


PROP.  II.     THEOR. 

If  any  two  points  be  taken  in  the  circumference  of  a  circle,  the  straight  line 
which  joins  them  shall  fall  within  the  circle. 

Let  ABC  be  a  circle,  and  A,  B  any  two  points  in  the  circumference  ; 
the  straight  line  drawn  from  A  to  B  shall  fall 
within  the  circle. 

Take  any  point  in  AB  as  E  ;  find  D  (1.  3.) 
the  centre  of  the  circle  ABC ;  join  AD,  DB 
and  DE,  and  let  DE  meet  the  circumference 
in  F.  Then,  because  DA  is  equal  to  DB,  the 
^ngle  DAB  is  equal  (5.  1.)  to  the  angle  DBA  ; 
and  because  AE,  a  side  of  the  triangle  DAE, 
is  produced  to  B,  the  angle  DEB  is  greater 
(16.  1.)  than  the  angle  DAE  ;  but  DAE  is 
equal  to  the  angle  DBE  ;  therefore  the  angle  DEB  is  greater  than  the 
angle  DBE:  Now  to  the  greater  angle  the  greater  side  is  opposite  (19. 
1.) ;  DB  is  therefore  greater  than  DE  :  but  BD  is  equal  to  DF ;  where- 
fore DF  is  greater  than  DE,  and  the  point  E  is  therefore  within  the  circle. 
The  same  may  be  demonstrated  of  any  other  point  between  A  and  B, 
therefore  AB  is  within  the  circle. 

CoR.  Every  point,  moreover,  in  the  production  of  AB,  is  farther  from  the 
centre  than  the  circumference. 

PROP.  III.     THEOR. 

If  a  straight  line  drawn  through  the  centre  of  a  circle  bisect  a  straight  line  in 
the  circle,  which  does  not  pass  through  the  centre,  it  will  cut  that  line  at  right 
angles  ;  and  if  it  cut  it  at  right  angles,  it  will  bisect  it. 

Let  ABC  be  a  circle,  and  let  CD,  a  straight  line  drawn  through  the 
centre,  bisect  any  straight  line  AB,  which  does  not  pass  through  the 
centre,  in  the  point  F  ;  it  cuts  it  also  at  right  angles. 

Take  (1.  3.)  E  the  centre  of  the  circle,  and  join  EA,  EB.  Then  be- 
cause AF  is  equal  to  FB,  andFE  common  to  the 
two  triangles  AFE,  BFE,  there  are  two  sides  in  the 
one  equal  to  two  sides  in  the  other :  but  the  base 
EA  is  equal  to  the  base  EB  ;  therefore  the  angle 
AFE  is  equal  (8.  1.)  to  the  angle  BFE.  And 
when  a  straight  line  standing  upon  another  makes 
the  adjacent  angles  equal  to  one  another,  each  of 
them  is  a  right  (7.  Def.  1.)  angle  :  Therefore  each 
of  the  angles  AFE,  BFE  is  a  right  angle  ;  where- 
fore the  straight  line  CD,  drawn  through  the  centre 


64  ELEMENTS 

bisecting  AB,  which  does  not  pass  through  the  centre,  cuts  AB  at  right 
angles. 

Again,  let  CD  cut  AB  at  right  angles  ;  CD  also  bisects  AB,  that  is,  AF 
is  equal  to  FB. 

The  same  construction  being  made,  because  the  radii  EA,  EB  are  equal 
to  one  another,  the  angle  EAF  is  equal  (5.  1.)  to  the  angle  EBF ;  and 
the  right  angle  AFE  is  equal  to  the  right  angle  BFE  :  Therefore,  in  the 
two  triangles  EAF,  EBF,  there  are  two  angles  in  one  equal  to  two  angles 
in  the  other ;  now  the  side  EF,  which  is  opposite  to  one  of  the  equal  an- 
gles in  each,  is  common  to  both ;  therefore  the  other  sides  are  equal  to 
(28.  1.) :  AF  therefore  is  equal  to  FB. 

Cor.  1 .  Hence,  the  perpendicular  through  the  middle  of  a  chord^  passes 
through  the  centre ;  for  this  perpendicular  is  the  same  as  the  one  let  fall 
from  the  centre  on  the  same  chord,  since  both  of  them  passes  through  the 
middle  of  the  chord. 

Cor.  2.  It  likewise  follows,  that  the  perpendicular  drawn  through  the 
middle  of  a  chord,  and  terminated  both  ways  by  the  circumference  of  the  circle, 
is  a  diameter^  and  the  middle  point  of  that  diameter  is  therefore  the  centre  of 
the  circle. 

PROP.  IV.    THEOR. 

If  in  a  circle  two  straight  lines  cut  one  another,  which  do  not  both  pass  through 
the  centre,  they  do  not  bisect  each  other. 

Let  ABCD  be  a  circle,  and  AC,  BD  two  straight  lines  in  it,  which  cut 
one  another  in  the  point  E,  and  do  not  both  pass  through  the  centre  :  AC, 
BD  do  not  bisect  one  another. 

For  if  it  is  possible,  let  AE  be  equal  to  EC,  and  BE  to  ED ;  if  one  of  the 
lines  pass  through  the  centre,  it  is  plain  that  it 
cannot  be  bisected  by  the  other,  which  does  not 
pass  through  the  centre.  But  if  neither  of  them 
pass  through  the  centre,  take  (1.  3.)  F  the  centre 
of  the  circle,  and  join  EF :  and  because  FE,  a 
straight  line  through  the  centre,  bisects  another 
AC,  which  does  not  pass  through  the  centre,  it 
must  cut  it  at  right  (3.  3.)  angles ;  wherefore 
FEA  is  a  right  angle.  Again,  because  the 
straight  line  FE  bisects  the  straight  line  BD,  which  does  not  pass  throuoh 
the  centre,  it  must  cut  it  at  right  (3.  3.)  angles  ;  wherefore  FEB  is  a  right 
angle  :  and  FEA  was  shown  to  be  a  right  angle  :  therefore  FEA  is  eoual 
to  the  angle  FEB,  the  less  to  the  greater,  which  is  impossible  ;  therefore 
AC,  BD,  do  not  bisect  one  another. 

PROP.  V.    THEOR. 

If  two  circles  cut  one  another,  they  cannot  have  the  same  centre. 

Let  the  two  circles  ABC,  CDG  cut  one  another  in  the  points  B,  C  ; 
they  have  not  the  same  centre. 


OF  GEOMETRY.    BOOK  III. 


65 


For,  if  it  be  possible,  let  E  be  their 
centre  :  join  EC,  and  draw  any  straight  line 
EFG  meeting  the  circles  in  F  |nd  G :  and 
because  E  is  the  centre  of  the  circle  ABC, 
CE  is  equal  to  EF :  Again,  because  E  is 
the  centre  of  the  circle  €DG,  CE  is  equal  to 
EG  :  but  CE  was  shown  to  be  equal  to  EF, 
therefore  EF  is  equal  to  EG,  the  less  to  the 
greater,  which  is  impossible  :  therefore  E 
is  not  the  centre  of  the  circles,  ABC,  CDG. 


PROP.  VI.     THEOR. 

If  two  circles  touch  one  another  internally,  they  cannot  have  the  same  centre. 

Let  the  two  circles  ABO,  CDE,  touch  one  another  internally  in  the 
point  C  ;  they  have  not  the  same  centre. 

*  For, 'if  they  have,  let  it  be  F  ;  join  FC,  and 
draw  any  straight  line  FEB  meeting  the  circles 
in  E  and  B  ;  and  because  F  is  the  centre  of 
the  circle  ABC,  CF  is  equal  to  FB  ;  also,  be- 
cause F  is  the  centre  of  the  circle  CDE,  CF 
is  equal  to  FE  :  but  CF  was  shown  to  be  equal 
to  FB  ;  therefore  FE  is  equal  to  FB,  the  less 
to  the  greuter,  which  is  impossible  ;  Where- 
fore F  is  not  the  centre  of  the  circles  ABC, 
CDE. 

PROP.  VII.     THEOR. 


If  any  point  he  taken  in  the  diameter  of  a  circle  which  is  not  the  centre,  of  all 
the  straight  lines  which  can  he  drawn  from  it  to  the  circumference,  the  great- 
est is  that  in  which  the  centre  is,  and  the  other  part  of  that  diameter  is  the 
least ;  and,  of  any  others,  that  which  is  nearer  to  the  line  passing  through 
the  centre  is  always  greater  than  one  more  remote  from  it ;  And  from  the 
same  point  there  can  he  drawn  only  two  straight  lines  that  are  equal  to  one 
another,  one  upon  each  side  of  the  shortest  line. 

Let  ABCD  be  a  circle,  and  AD  its  diameter,  in  which  let  any  point  F 
be  taken  which  is  not  the  centre  :  let  the  centre  be  E  ;  of  all  the  straight 
lines  FB,  FC,  FG,  &c.  that  can  be  drawn  from  F  to  the  circumference, 
FA  is  the  greatest  ;  and  FD,  the  other  part  of  the  diameter  AD,  is  the 
least ;  and  of  the  others,  FB  is  greater  than  FC,  and  FC  than  FG. 

Join  BE,  CE,  GE  ;  and  because  two  sides  of  a  triangle  are  greater 
(20.  1.)  than  the  third,  BE,  EF  are  greater  than  BF  ;  but  AE  is  equal  to 
EB  ;  therefore  AE  and  EF,  that  is,  AF,  is  greater  than  BF  :  Again,  be- 
cause BE  is  equal  to  CE,  and  FE  common  to  the  triangles  BEF,  CEF, 

9 


66  ELEMENTS 

the  two  sides  BE,  EF  are  equal  to  the  two 
CE,  EF;  but  the  angle  BEF  is  greater  than 
the  angle  CEF  ;  therefore  the  base  Bf  is 
greater  (24, 1.)  than  the  base  FC ;  for  the  same 
reason,  CF  is  greater  than  GF.  Again,  be- 
cause GF,  FE  are  greater  (20.  1.)  than  EG, 
and  EG  is  equal  to  ED  ;  GF,  FE  are  greater 
than  ED  ;  take  away  the  common  part  FE, 
and  the  remainder  GF  is  greater  than  the  re- 
mainder FD :  therefore  FA  is  the  greatest,  and 
FD  the  least  of  all  the  straight  lines  from  F  to 
the  circumference  ;  and  BF  is  greater  than  GF,  and  CF  than  GF. 

Also  there  can  be  drawn  only  two  equal  straight  lines  from  the  point  F 
to  the  circumference,  one  upon  each  side  of  the  shortest  line  FD  :  at  the 
point  E  in  the  straight  line  EF,  make  (23.  1.)  the  angle  FEH  equal  to  the 
angle  GEF,  and  join  FH  :  Then,  because  GE  is  equal  to  EH,  and  EF  com- 
mon to  the  two  triangles  GEF,  HEF  ;  the  two  sides  GE,  EF  are  equal 
to  the  two  HE,  EF ;  and  the  angle  GEF  is  equal  to  the  angle  HEF  ; 
therefore  the  base  FG  is  equal  (4.  1.)  to  the  base  FH  :  but  besides  FH, 
no  straight  line  can  be  drawn  from  F  to  the  circumference  equal  to 
FG  :  for,  if  there  can,  let  it  be  FK  ;  and  because  FK  is  equal  to  FG,  and 
FG  to  FH,  FK  is  equal  to  FH  ;  that  is,  a  line  nearer  to  that  which  passes 
through  the  centre,  is  equal  to  one  more  remote,  which  is  impossible. 

PROP.  Vni.    THEOR. 

If  any  point  he  taken  without  a  circle^  and  straight  lines  be  drawn  from  it  to 
the  circumference,  whereof  one  passes  through  the  centre  ;  of  those  which 
fall  upon  the  concave  circumference,  the  greatest  is  that  which  passes  through 
the  centre ;  and  of  the  rest  that  which  is  nearer  to  that  through  the  centre 
is  always  greater  than  the  more  remote  ;  But  of  those  which  fall  upon  the 
convex  circumference,  the  least  is  that  between  the  point  without  the  circle, 
and  the  diameter  ;  and  of  the  rest,  that  lohich  is  nearer  to  the  least  is  al- 
ways less  than  the  more  remote  :  And  only  two  equal  straight  lines  can  be 
drawn  from  the  point  unto  the  circumference,  one  upon  each  side  of  the  least. 

Let  ABC  be  a  circle,  and  D  any  point  without  it,  from  which  let  the 
straight  lines  DA,  DE,  DF,  DC  be  drawn  to  the  circumference,  whereof  DA 
passes  through  the  centre.  Of  those  which  fall  upon  the  concave  part  of  the 
circumference  AEFC,  the  greatest  is  AD,  which  passes  through  the  cen- 
tre ;  and  the  line  nearer  to  AD  is  always  greater  than  the  more  remote, 
viz.  DE  than  DF,  and  DF  than  DC  ;  but  of  those  which  fall  upon  the  con- 
vex circumference  HLKG,  the  least  is  DG,  between  the  point  D  and  the 
diameter  AG  ;  and  the  nearer  to  it  is  always  less  than  the  more  remote, 
viz.  DK  than  DL,  and  DL  than  DH. 

Take  (1.  3.)  M  the  centre  of  the  circle  ABC,  and  join  ME,  MF,  MC, 
MK,  ML,  MH  :  And  because  AM  is  equal  to  ME,  if  MD  be  added  to 
each,  AD  is  equal  to  EM  and  MD  ;  but  EM  and  MD  are  greater  (20.  1.) 
than  ED  :  therefore  also  AD  is  greater  than  ED.  Again,  because  ME  is 
equal  to  MF,  and  MD  common  to  the  triangles  EMD,  FMD ;  EM,  MD 


OF  GEOMETRY.     BOOK  III. 


are  equal  to  FxM,  MD  ;  but  the  angle  EMD  is  greater  than  the  angle 

FMD  ;  therefore  the  base  ED  is  greater 

(24.  1.)  than  the  base  FD.    In  like  manner 

it  may  be  shewn  that  FD  is  greater  than 

CD.     Therefore  DA  is  the  greatest ;  and 

DE  greater  than  DF,  and  DF  than  DC. 

And  because  MK,  KD  are  greater  (20. 
1.)  than  MD,  and  MK  is  equal  to  MG,  the 
remainder  KD  is  greater  (5.  Ax.)  than  the 
remainder  GD,  that  is,  GD  is  less  than 
KD  :  And  because  MK,  DK  are  drawn  to 
the  point  K  within  the  triangle  MLD  from 
M,  D,  the  extremities  of  its  side  MD  ;  MK, 
KD  are  less  (21.1.)  than  ML,  LD,  whereof 
MK  is  equal  to  ML  ;  therefore  the  remain- 
der DK  is  less  than  the  remainder  DL  : 
In  like  manner,  it  may  be  shewn  that  DL 
is  less  than  DH  :  Therefore  DG  is  the 
least,  and  DK  less  than  DL,  and  DL 
than  DH. 

Also  there  can  be  drawn  only  two  equal  straight  lines  from  the  point  D 
to  the  circumference,  one  upon  each  side  of  the  least ;  at  the  point  M,  in 
the  straight  line  MD,  make  the  angle  DMB  equal  to  the  angle  DMK,  and 
join  DB  ;  and  because  in  the  triangles  KMD,  BMD,  the  side  KM  is  equal 
to  the  side  BM,  and  MD  common  to  both,  and  also  the  angle  KMD  equal 
to  the  angle  BMD,  the  base  DK  is  equal  (4.  l.)to  the  base  DB.  But, 
besides  DB,no  straight  line  can  be  drawn  from  D  to  the  circumference,  equal 
to  DK  ;  for,  if  there  can,  let  it  be  DN  ;  then,  because  DN  is  equal  to  DK, 
and  DK  equal  to  DB,  DB  is  equal  to  DN  ;  that  is,  the  line  nearer  to  DG, 
the  least,  equal  to  the  more  remote,  which  has  been  shewn  to  be  impossible. 

PROP.  IX.    THEOR. 

If  a  point  he  taken  within  a  circle,  from  which  there  fall  more  than  two  equal 
straight  lines  upon  the  circumference,  that  point  is  the  centre  of  the  circle. 

Let  the  point  D  be  taken  within  the  circle  ABC,  from  which  there  fall 
on  the  circumference  more  than  two  equal  straight  lines,  viz.  DA,  DB,  DC, 
the  point  D  is  the  centre  of  the  circle. 

For,  if  not,  let  E  be  the  centre,  join  DE,  and  produce  it  to  the  circum- 
ference in  F,  G ;  then  FG  is  a  diameter  of 
the  circle  ABC  :  And  because  in  FG,  the  di- 
ameter of  the  circle  ABC,  there  is  taken  the 
point  D  which  is  not  the  centre,  DG  is  the 
greatest  line  from  it  to  the  circumference,  and 
DC  greater  (7.  3.)  than  DB,  and  DB  than 
DA ;  but  they  are  likewise  equal,  which  is 
impossible  :  Therefore  E  is  not  the  centre  of 
the  circle  ABC :  In  like  manner  it  may  be 
demonstrated,  that  no  other  point  but  D  is  the 
centre :  D  therefore  is  the  centre. 


n 


ELEMENTS 


PROP.  X.    THEOR. 


One  circle  cannot  cut  another  in  more  than  two  points. 

If  It  be  possible,  let  the  circumference  FAB  cut  the  circumference  DEF 
in, more  than  two  points,  viz.  in  B,  G,  F ;  take  the  centre  K  of  the  circle 
ABC,  and  join  KB,  KG,  KF  ;  and  because  within  the  circle  DEF  there 
is  taken  the  point  K,  from  which  more  than  two 
equal  straight  lines,  viz.  KB,  KG,  KF,  fall  on 
the  circumference  DEF,  the  point  K  is  (9.  3.) 
the  centre  of  the  circle  DEF  ;  but  K  is  also  the 
centre  of  the  circle  ABC  ;  therefore  the  same 
point  is  the  centre  of  two  circles  that  cut  one 
another,  which  is  impossible  (5.  3.).  There- 
fore one  circumference  of  a  circle  cannot  cut 
another  in  more  than  two  points. 


PROP.  XI.     THEOR. 


I 


If  two  circles  touch  each  other  internally,  the  straight  line  which  joins  their 
centres  being  produced,  will  pass  through  the  point  of  contact. 

Let  the  two  circles  ABC,  ADE,  touch  each  other  internally  in  the  point 
A,  and  let  F  be  the  centre  of  the  circle  ABC,  and  G  the  centre  of  the  cir- 
cle ADE  ;  the  straight  line  which  joins  the  cen- 
tres F,  G,  being  produced,  passes  through  the 
point  A. 

For,  if  not,  let  it  fall  otherwise,  if  possible,  as 
FGDH,  and  join  AF,  AG :  And  because  AG, 
GF  are  greater  (20.  1.)  than  FA,  that  is,  than 
FH,  for  FA  is  equal  to  FH,  being  radii  of  the 
same  circle ;  take  away  the  common  part  FG, 
and  the  remainder  AG  is  greater  than  the  re- 
mainder GH.  But  AG  is  equal  to  GD,  there- 
fore GD  is  greater  than  GH  ;  and  it  is  also  less, 
which  is  impossible.  Therefore  the  straight  line 
which  joins  the  points  F  and  G  cannot  fall  otherwise  than  on  the  point 
A  ;  that  is,  it  must  pass  through  A. 

Cor.  1.  If  two  circles  touch  each  other  internally,  the  distance  be- 
tween their  centre  must  be  equal  to  the  difference  of  their  radii :  for  the 
circumferences  pass  through  the  same  point  in  the  line  joining  the  centres. 

CoR.  2.  And,  conversely,  if  the  distance  between  the  centres  be  equa) 
to  the  difference  of  the  radii,  the  two  circles  will  touch  each  other  inters 
nally. 


OF  GEOMETRY.    BOOK  III. 


69 


PROP.  XII.     THEOR. 

If  two  circles  touch  each  other  externally,  the  straight  line  which  joins  their 
centres  will  pass  through  the  point  of  contact. 

Let  the  two  circles  ABC,  ADE,  touch  each  other  externally  in  the  point 
A ;  and  let  F  be  the  centre  of  the  circle  ABC,  and  G  the  centre  of  ADE  ; 
the  straight  line  which  joins  the  points  F,  G  shall  pass  through  the  point 
of  contact. 

For,  if  not,  let  it  pass  otherwise,  if  possible,  FCDG,  and  join  FA,  AG  : 
and  because  F  is  the  centre  of  the  circle  ABC,  AF  is  equal  to  FC  :  Also 
because  G  is  the  centre  of  the 
circle,  ADE,  AG  is  equal  to 
GD.  Therefore  FA,  AG  are 
equal  to  FC,  DG  ;  wherefore 
the  whole  FG  is  greater  than 
FA,  AG ;  but  i^  is  also  less 
(20.  1 .),  which  is  impossible  : 
Therefore  the  straight  line 
which  joins  the  points  F,  G 
cannot  pass  otherwise  than 
through  the  point  of  contact  A  ;  that  is,  it  passes  through  A. 

CoR.  Hence,  if  two  circles  touch  each  other  externally,  the  distance 
between  their  centres  will  be  equal  to  the  sum  of  their  radii. 

And,  conversely,  if  the  distance  between  the  centres  be  equal  to  the  sum 
of  the  radii,  the  two  circles  will  touch  each  other  externally. 

PROP.  XIII.     THEOR. 

One  circle  cannot  touch  another  in  more  points  than  one,  whether  it  touches 
it  on  the  inside  or  outside. 

For,  if  it  be  possible,  let  the  circle  EBF  touch  the  circle  ABC  in  more 
points  than  one,  and  first  on  the  inside,  in  the  points  B,  D  ;  join  BD,  and 
draw  (10.  11.  1.)  GH,  bisecting  BD  at  right  angles  :  Therefore  because 
*he  points  B,  D  are  in  the  circumference  of  each  of  the  circles,  the  straio-ht 


D     G 


line  BD  falls  within  each  (2.  3.)  of  them  :  and  therefore  their  centres  are 
(Cor.  1.  3.)  in  the  straight  line  GH  which  bisects  BD  at  right  angles : 


70 


ELEMENTS 


therefore  GH  passes  through  the  point  of  contact  (11.  3.);  but  it  does 
not  pass  through  it,  because  the  points  B,  D  are  without  the  straight  line 
GH,  which  is  absurd :  therefore  one  circle  cannot  touch  another  in  the 
inside  in  more  points  than  one. 

Nor  can  two  circles  touch  one  another  on  the  outside  in  more  than  one 
point :  For,  if  it  be  possible,  let  the  circle  ACK 
touch  the  circle  ABC  in  the  points  A,  0,  and  join 
AC  :  therefore,  because  the  two  points  A,  C  are 
in  the  circumference  of  the  circle  ACK,  the  straight 
line  AC  which  joins  them  shall  fall  within  the 
circle  ACK :  And  the  circle  ACK  is  without  the 
circle  ABC  :  and  therefore  the  straight  line  AC  is 
also  without  ABC ;  but,  because  the  points  A,  C 
are  in  the  circumference  of  the  circle  ABC,  the 
straight  line  AC  must  be  within  (2.  3.)  the  same 
circle,  which  is  absurd :  therefore  a  circle  cannot 
touch  another  on  the  outside  in  more  than  one 
point :  and  it  has  been  shewn,  that  a  circle  cannot 
touch  another  on  the  inside  in  more  than  one  point. 


PROP.  XIV.     THEOR. 


Equal  straight  lines  in  a  circle  are  equally  distant  from  the  centre ;  and  those 
which  are  equally  distant  from  the  centre,  are  equal  to  ona  another. 

Let  the  straight  lines  AB,  CD,  in  the  circle  ABDC,  be  equal  to  one 
another :  they  are  equally  distant  from  the  centre. 

Take  E  the  centre  of  the  circle  ABDC,  and  from  it  draw  EF,  EG,  per- 
pendiculars to  AB,  CD  ;  join  AE  and  EC.  Then,  because  the  straight 
line  EF  passing  through  the  centre,  cuts  the 
straight  line  AB,  which  does  not  pass  through 
the  centre  at  right  angles,  it  also  bisects  (3. 
3.)  it :  Wherefore  AF  is  equal  to  FB,  and 
AB  double  of  AF.  For  the  same  reason, 
CD  is  double  of  CG  :  But  AB  is  equal  to 
CD  ;  therefore  AF  is  equal  to  CG  :  And  be- 
cause AE  is  equal  toEC,  the  square  of  AE  is 
equal  to  the  square  of  EC  :  Now  the  squares 
of  AF,  FE  are  equal  (47.  1.)  to  the  square 
of  AE,  because  the  angle  AFE  is  a  right  angle  ;  and,  for  the  like  reason, 
the  squares  of  EG,  GC  are  equal  to  the  square  of  EC  :  therefore  the 
squares  of  AF,  FE  are  equal  to  the  squares  of  CG,  GE,  of  w^hich  the 
square  of  AF  is  equal  to  the  square  of  CG,  because  AF  is  equal  to  CG  ; 
therefore  the  remaining  square  of  FE  is  equal  to  the  remaining  square  of 
EG,  and  the  straight  line  EF  is  therefore  equal  to  EG  :  But  straight  lines 
in  a  circle  are  said  to  be  equally  distant  from  the  centre  when  the  perpen- 
diculars drawn  to  them  from  the  centre  are  equal  (3.  Def.  3.) :  therefore 
AB,  CD  are  equally  distant  from  the  centre. 

Next,  if  the  straight  lines  AB,  CD  be  equally  distant  from  the  centre, 
that  is,  if  FE  be  equal  to  EG,  AB  is  equal  to  CD.     For,  the  same  con- 


/ 
/ 


OF  GEOMETRY.     BOOK  III.  71 

struction  being  made,  it  may,  as  before,  be  demonstrated,  that  AB  is  double 
of  AF,  and  CD  double  of  CG,  and  that  the  squares  of  EF,  FA  are  equal 
to  the  squares  of  EG,  GO  ;  of  which  the  square  of  FE  is  equal  to  the 
square  of  EG,  because  FE  is  equal  to  EG  :  .therefore  the  remaining  square 
of  AF  is  equal  to  the  remaining  square  of  CG  ;  and  the  straight  line  AF 
is  therefore  equal  to  CG  :  But  AB  is  double  of  AF,  and  CD  double  of 
CG  ;  wherefore  AB  is  equal  to  CD.  * 

PROP.  XV.     THEOR. 

The  diameter  is  the  greatest  straight  line  in  a  circle ;  and  of  all  others, 
that  which  is  nearer  to  the  centre  is  always  greater  than  one  more  remote  ; 
and  the  greater  is  nearer  to  the  centre  than  the  less. 

Let  ABCD  be  a  circle,  of  which  the  diame- 
ter is  AD,  and  the  centre  E  ;  and  let  BC  be  near- 
er to  the  centre  than  FG  ;  AD  is  greater  than 
any  straight  line  BC  which  is  not  a  diameter,  and 
BC  greater  than  FG. 

From  the  centre  draw  EH,EK  perpendiculars 
to  BC,  FG,  and  join  EB,  EC,  EF  ;  and  because 
AE  is  equal  to  EB,  and  ED  to  EC,  AD  is  equal 
to  EB,  EC  :  But  EB,  EC  are  greater  (20.  1.) 
than  BC ;  wherefore,  also,  AD  is  greater  than 
BC. 

And,  because  BC  is  nearer  to  the  centre  than  FG,  EH  is  less  (4.  Def. 
3.)  than  EK ;  But,  as  was  demonstrated  in  the  preceding,  BC  is  double 
of  BH,  and  FG  double  of  FK,  and  the  squares  of  EH,  HB  are  equal  to 
the  squares  of  EK,  KF,  of  which  the  square  of  EH  is  less  than  the  square 
of  EK,  because  EH  is  less  than  EK  ;  therefore  the  square  of  BH  is  greater 
than  the  square  of  FK,  and  the  straight  line  BH  greater  than  FK  :  and 
therefore  BC  is  greater  than  FG. 

Next,  let  BC  be  greater  than  FG  ;  BC  is  nearer  to  the  centre  than  FG  : 
that  is,  the  same  construction  being  made,  EH  is  less  than  EK  ;  because 
BC  is  greater  than  FG,  BH  likewise  is  greater  than  KF  :  but  the  squares 
of  BH,  HE  are  equal  to  the  squares  of  FK,  KE,  of  which  the  square  of 
BH  is  greater  than  the  square  of  FK,  because  BH  is  greater  than  FK  ; 
therefore  the  square  of  EH  is  less  than  the  square  of  EK,  and  the  straight 
line  EH  less  than  EK. 

CoR.  The  shorter  the  chord  is,  the  farther  it  is  from  the  centre  ;  and, 
conversely,  the  farther  the  chord  is  from  the  centre,  the  shorter  it  is. 

PROP.  XVI.     THEOR. 

The  straight  line  drawn  at  right  angles  to  the  diameter  of  a  circle,  from  the 
extremity  of  it,  falls  without  the  circle ;  and  no  straight  line  can  he  drawn 
between  that  straight  line  and  the  circumference,  from  the  extremity  of  the 
diameter,  so  as  not  to  cut  the  circle. 

Let  ABC  be  a  circle,  the  centre  of  which  is  D,  and  the  diameter  AB  : 
and  let  AE  be  drawn  from  A  perpendicular  to  AB,  AE  shall  fall  without 
the  circle. 


m 


72 


ELEMENTS 


In  AE  take  any  point  F,  join  DF  and  let  DF  meet  the  circle  in  C. 
Because  DAF  is  a  right  angle,  it  is  greater 
than  the  angle  AFD  (32. 1.) ;  but  the  greater 
angle  of  any  triangle  is  subtended  by  the 
greater  side  (19.  1.),  therefore  DF  is  greater 
than  DA  :  noAV  T)A  is  equal  to  DC,  there- 
fore DF  is  greater  than  DC,  and  the  point 
F  is  therefore  without  the  circle.  And  F 
is  any  point  whatever  in  the  line  AE,  there- 
fore AE  falls  without  the  circle. 

Again,  between  the  straight  line  AE  and 
the  circumference,  no  straight  line  can  be 
drawn  from  the  point  A,  which  does  not  cut 

the  circle.  Let  AG  be  drawn  in  the  angle  DAE  :  from  D  draw  DH  at 
right  angles  to  AG ;  and  because  the  angle 
DHA  is  a  right  angle,  and  the  angle  DAH 
less  than  a  right  angle,  the  side  DH  of  the 
triangle  DAH  is  less  than  the  side  DA  (19. 
1 .).  The  point  H,  therefore,  is  within  the  cir- 
cle, and  therefore  the  straight  line  AG  cuts 
the  circle. 

CoR.  1.  From  this  it  is  manifest,  that  the  -g 
straight  line  which  is  drawn  at  right  angles  to 
the  diameter  of  a  circle  from  the  extremity  of 
it,  touches  the  circle ;  and  that  it  touches  it 
only  in  one  point ;  because,  if  it  did  meet  the 
circle  in  two,  it  would  fall  within  it  (2.  3.). 
Also  it  is  evident  that  there  can  be  but  one  straight  line  which  touches  the 
circle  in  the  same  point. 

CoR.  2.  Hence,  a  perpendicular  at  the  extremity  of  a  diameter  is  a  tan- 
gent to  the  circle  ;  and,  conversely,  a  tangent  to  a  circle  is  perpendicular 
to  the  diameter  drawn  from  the  point  of  contact. 

CoR.  3.  It  follows,  likewise,  that  tangents  at  each  extremity  of  the 
diameter  are  parallel  (Cor.  28.  B.  1.);  and,  conversely,  parallel  tangents 
are  both  perpendicular  to  the  same  diameter,  and  have  their  points  of  con- 
tact at  its  extremities. 


PROP.  XVII.     PROB. 


To  draw  a  straight  line  from  a  given  point  either  without  or  in  the  circum' 
fercncc^  which  shall  touch  a  given  circle. 

First,  let  A  be  a  given  point  without  the  given  circle  BCD ;  it  is  re- 
quired to  draw  a  straight  line  from  A  which  shall  touch  the  circle. 

Find  (1.3.)  the  centre  E  of  the  circle,  and  join  AE  ;  and  from  the  cen- 
tre E,  at  the  distance  EA,  describe  the  circle  AFG ;  from  the  point  D 
draw  (11.  1.)  DF  at  right  angles  to  E A,  join  EBF,  and  draw  AB.  AB 
touches  the  circle  BCD. 

Because  E  is  the  centre  of  the  circles  BCD,  AFG,  EA  is  equal  to 
EF,  and  EP  to  EB ;  therefore  the  two  sides  AE  EB  are  equal  to  the 


OF  GEOMETRY.    BOOK  III. 


7k. 


two  FE,  ED,  and  they  contain  the  angle  at  E  common  to  the  two  trian- 
gles AEB,  FED ;  therefore  the  base  DF 
is  equal  to  the  base  AB,  and  the  triangle 
FED  to  the  triangle  AEB,  and  the  other 
angles  to  the  other  angles  (4.  1.)  ;  there-  G/ 
fore  the  angle  EBA  is  equal  to  the  angle 
EDF;  but  EDF  is  a  right  angle,  where- 
fore EBA  is  a  right  angle;  and  EB  is  a 
line  drawn  from  the  centre  :  but  a  straight 
line  drawn  from  the  extremity  of  a  diame- 
ter, at  right  angles  to  it,  touches. the  circle 
(1  Cor.  16.3.):  therefore  AB  touches  the 
circle  ;  and  is  drawn  from  the  given  point  A. 

But  if  the  given  point  be  in  the  circumference  of  the  circle,  as  the  point 
D,  draw  DE  to  the  centre  E,  and  DF  at  right  angles  to  DE  ;  DF  touches 
the  circle  (1  Cor.  16.  3 1) 

SCHOLIUM. 

When  the  point  A  lies  without  the  circle,  there  will  evidently  be  always 
two  equal  tangents  passing  through  the  point  A.  For,  by  producing  the 
tangent  FD  till  it  meets  the  circumference  AG,  and  joining  E  and  the  point 
of  intersection,  and  also  A  and  the  point  where  this  last  line  will  intersect 
the  circumference  DC  ;  there  will  be  formed  a  right  angled  triangle  equal 
to  ABE  (46.  1.). 

PROP.  XVIII.    THEOR. 

If  a  straight  line  touch  a  circle^  the  straight  line  drawn  from  the  centre  to 
the  point  of  contact ^  is  perpendicular  to  the  line  touching  the  circle. 


Let  the  straight  line  DE  touch  the  circle  ABC  in  the  point  C  ;  take 
the  centre  F,  and  draw  the  straight  line  FC  :  FC  is  perpendicular  to  DE. 

For,  if  it  be  not,  from  the  point  F  draw  FBG  perpendicular  to  DE  ;  and 
because  FGC  is  a  right  angle,  GCF  must  * 

be  (17.  1.)  an  acute  angle  ;  and  to  the  great- 
er angle  the  greater  side  (19.  1.)  is  oppo- 
site ;  therefore  FC  is  greater  than  FG ; 
but  FC  is  equal  to  FB  ;  therefore  FB  is 
greater  than  FG,  the  less  than  the  greater, 
which  is  impossible  ;  wherefore  FG  is  not 
perpendicular  to  DE  :  in  the  same  manner 
it  may  be  shewn,  that  no  other  line  but  FC 
can  be  perpendicular  to  DE  ;  FC  is  there- 
fore perpendicular  to  DE. 


10 


O     E 


74 


ELEMENTS 


PROP.  XIX.    THEOR. 

If  a  straight  line  touch  a  circle,  and  from  the  point  of  contact  a  straight  line 
be  drawn  at  right  angles  to  the  touching  line,  the  centre  of  the  circle  is  in 
that  line. 

Let  the  straight  line  DE  touch  the  circle  ABC,  in  C,  and  from  C  let    ^ 
CA  be  drawn  at  right  angles  to  DE  ;  the  centre  of  the  circle  is  in  CA. 

For,  if  not,  let  F  be  the  centre,  if  possible, 
and  join  CF.  Because  DE  touches  the  cir- 
cle ABC,  and  FC  is  drawn  from  the  centre 
to  the  point  of  contact,  FC  is  perpendicular 
(18.  3.)  to  DE  ;  therefore  FCE  is  a  right 
angle ;  but  ACE  is  also  a  right  angle ; 
therefore  the  angle  FCE  is  equal  to  the  an- 
gle ACE,  the  less  to  the  greater,  which  is 
impossible  ;  Wherefore  F  is  not  the  centre 
of  the  circle  ABC :  in  the  same  manner  it 
may  be  shewn,  that  no  other  point  which  is 
not  in  CA,  is  the  centre  ;  that  is,  the  centre 
is  in  CA. 

PROP.  XX.    THEOR. 

The  angle  at  the  centre  of  a  circle  is  double  of  the  angle  at  the  circumfer- 
ence, upon  the  same  base,  that  is,  upon  the  same  part  of  the  circumfer- 


Let  ABC  be  a  circle,  and  BDC  an  angle  at  the  centre,  and  BAC  an 
angle  at  the  circumference  which  have  the  same  circumference  BC  for 
the  base ;  the  angle  BDC  is  double  of  the  angle  BAC. 

First,  let  D,  the  centre  of  the  circle,  be  within  the  angle  BAC,  and  join 
AD,  and  produce  it  to  E  :  because  DA  is  equal 
to  DB,  the  angle  DAB  is  equal  (5.  1.)  to  the 
angle  DBA:  therefore  the  angles  DAB,  DBA 
together  are  double  of  the  angle  DAB  ;  but  the 
angle  BDE  is  equal  (32.  1.)  to  the  angles  DAB, 
DBA;  therefore  also  the  angle  BDE  is  double 
of  the  angle  DAB  ;  for  the  same  reason,  the  an- 
gle EDC  is  double  of  the  angle  DAC  :  there- 
fore the  whole  angle  BDC  is  double  of  the  whole 
angle  BAC. 

Again,  let  D,  the  centre  of  the  circle,  be 
without  the  angle  BAC ;  and  join  AD  and  pro- 
duce it  to  E.  It  may  be  demonstrated,  as  in 
the  first  case,  that  the  angle  EDC  is  double 
of  the  angle  DAC,  and  that  EDB,  a  part  of 
the  first,  is  double  of  DAB,  a  part  of  the 
other  ;  therefore  the  remaining  angle  BDC  is 
double  of  the  remaining  angle  BAC 


OF  GEOMETRY.    BOOK  III. 


75 


PROP.  XXI.     THEOR. 
The  angles  in  the  same  segment  of  a  circle  are  equal  to  one  another. 

Let  ABCD  be  a  circle,  and  BAD,  BED 
angles  in  the  same  segment  BAED  :  the  an- 
gles BAD,  BED  are  equal  to  one  another. 

Take  F  the  centre  of  the  circle  ABCD  : 
And,  first,  let  the  segment  BAED  be  greater 
than  a  semicircle,  and  join  BF,  FD :  and  be- 
cause the  angle  BFD  is  at  the  centre,  and  the 
angle  BAD  at  the  circumference,  both  having 
the  same  part  of  the  circumference,  viz.  BCD, 
for  their  base ;  therefore  the  angle  BFD  is 
double  (20.  3.)  of  the  angle  BAD  :  for  the 
same  reason,  the  angle  BFD' is  double  of  the 
angle  BED  :  therefore  the  angle  BAD  is  equal 
to  the  angle  BED. 

But,  if  the  segment  BAED  be  not  greater 
than  a  semicircle,  let  BAD,  BED  be  angles 
in  it ;  these  also  are  equal  to  one  another. 
Draw  AF  to  the  centre,  and  produce  to  C,  and 
join  CE  :  therefore  the  segment  BADC  is 
greater  than  a  semicircle  ;  and  the  angles  in 
it,  BAG,  BEC  are  equal,  by  the  first  case  : 
for  the  same  reason,  because  CBED  is  great- 
er than  a  semicircle,  the  angles  CAD,  CED 
are  equal ;  therefore  the  whole  angle  BAD  is 
equal  to  the  whole  angle  BED. 


PROP.  XXII.    THEOR. 

The  opposite  angles  of  any  quadrilateral  figure  described  in  a  circle^  are 
together  equal  to  two  right  angles. 


Let  ABCD  be  a  quadrilateral  figure  in  the  circle  ABCD 
its  opposite  angles  are  together  equal  to  two  right  angles. 

Join  AC,  BD.  The  angle  CAB  is  equal  (21.  3.) 
CDB,  because  they  are  in  the  same  segment 
BADC,  and  the  angle  ACB  is  equal  to  the  an- 
gle ADB,  because  they  are  in  the  same  seo-- 
ment  ADCB  ;  therefore  the  whole  angle  ADC 
is  equal  to  the  angles  CAB,  ACB  :  to  each  of 
these  equals  add  the  angle  ABC ;  and  the  an- 
gles ABC,  ADC,  are  Iqual  to  the  angles  ABC, 
CAB,  BCA.  But  ABC,  CAB,  BCA  are  equal 
to  two  right  angles  (32.  1.) ;  therefore  also  the 
angles  ABC,  ADC  are  equal  to  two  right  an- 
gles ;  in  the  same  manner,  the  angles  BAD, 
DCB  may  be  shewn  to  be  equal  to  two  right  angles. 


any 


two  of 


to  the   angle 


76  *  ELEMENTS 

Cor.  1.  If  any  side  of  a  quadrilateral  be  produced,  the  exterior  angle 
will  be  equal  to  the  interior  opposite  angle.  •>    r*     { 

CoR.  2.  It  follows,  likewise,  that  a  quadrilateral,  of  which  the  op- 
posite angles  are  not  equal  to  two  right  angles,  cannot  be  inscribed  in  a 
circle. 

PROP.  XXIII.    THEOR. 

Upon  the  same  straight  line,  and  upon  the  same  side  of  it,  there  cannot  be 
two  similar  segments  of  circles,  not  coinciding  with  one  another. 

If  it  be  possible,  let  the  two  similar  segments  of  circles,  viz.  ACB,  ADB, 
be  upon  the  same  side  of  the  same  straight  line  AB,  not  coinciding  with 
one  another ;  then,  because  the  circles  ACB,  ADB,  cut  one  another  in 
the  two  points  A,  B,  they  cannot  cut  one  another  in  any  other  point  (10. 
3.) :  one  of  the  segments  must  therefore  fall 
within  the  other:  let  ACB  fall  within  ADB, 
draw  the  straight  line  BCD,  and  join  CA,  DA  : 
and  because  the  segment  ACB  is  similar  to  the 
segment  ADB,  and  similar  segments  of  circles 
contain  (9.  def.  3.)  equal  angles,  the  angle 
ACB  is  equal  to  the  angle  ADB,  the  exterior 
to  the  interior,  which  is  impossible  (16.  1.). 

PROP.  XXIV.    THEOR. 
Similar  segments  of  circles  upon  equal  straight  lines  are  equal  to  one  another. 

Let  AEB,  CFD  be  similar  segments  of  circles  upon  the  equal  straight 
lines  AB,  CD ;  the  segment  AEB  is  equal  to  the  segment  CFD. 

For,  if  the  segment  AEB  be  applied  to  the  segment  CFD,  so  as  the 
point  A  be  on  C,  and  the 
straight  line  AB  upon  CD, 
the  point  B  shall  coincide 
with  the  point  D,  because 

AB  is  equal  to  CD  :  there-  

fore  the  straight  line  AB       A.  J3    C  J) 

coinciding  with  CD,  the  segment  AEB  must  (23.  3.)  coincide  with  the 
segment  CFD,  and  therefore  is  equal  to  it. 

PROP.  XXV.    PROB. 

A  segment  of  a  circle  being  given,  to  describe  the  circle  of  which  it  is  the 

segment. 

Let  ABC  be  the  given  segment  of  a  circle  ;  it  is  required  to  describe 
the  circle  of  which  it  is  the  segment. 

Bisect  (10.  1.)  AC  in  D,  and  from  the  point  D  draw  (II.  1.)  DB  at 
right  angles  to  AC,  and  join  AB  :  First,  let  the  angles  ABD,  BAD  be 
equal  to  one  another;  then  the  straight  line  BD  is  equal  (6.  1.)  to  DA, 
and  therefore  to  DC  ;  and  because  the  three  straight  lines  DA,  DB,  DC, 


OF  GEOMETRY.    BOOK  III. 


77 


are  all  equal ;  D  is  the  centre  of  the  circle  (9.  3.) ;  from  the  centre  D,  at 
the  distance  of  any  of  the  three  DA,  DB,  DC,  describe  a  circle  ;  this  shall 
pass  through  the  other  points  ;  and  the  circle  of  which  ABC  is  a  segment 

B 


A,         D         V  E  AD         C 

is  described  :  and  because  the  centre  D  is  in  AC,  the  segment  ABC  is  a 
semicircle.  Next,  let  the  angles  ABD,  BAD  be  unequal ;  at  the  point  A,  in 
the  straight  line  AB,  make  (23.  1.)  the  angle  BAE  equal  to  the  angle  ABD, 
and  produce  BD,  if  necessary,  to  E,  and  join  EC  :  and  because  the  angle 
ABE  is  equal  to  the  angle  BAE,  the  straight  line  BE  is  equal  (6.  1.)  to 
EA  :  and  because  AD  is  equal  to  DC,  and  DE  common  to  the  triangles 
ADE,  CDE,  the  two  sides  AD,  DE  are  equal  to  the  two  CD,  DE,  each 
to  each  ;  and  the  angle  ADE  is  equal  to  the  angle  CDE,  for  each  of  them 
is  a  right  angle  ;  therefore  the  base  AE  is  equal  (4.  1.)  to  the  base  EC  : 
but  AE  was  shewn  to  be  equal  to  EB,  wherefore  also  BE  is  equal  to  EC  : 
and  the  three  straight  lines  AE,  EB,  EC  are  therefore  equal  to  one  another; 
wherefore  (9.  3.)  E  is  the  centre  of  the  circle.  From  the  centre  E,  at 
the  distance  of  any  of  the  three  AE,  EB,  EC,  describe  a  circle,  this  shall 
pass  through  the  other  points  ;  and  the  circle  of  which  ABC  is  a  segment 
is  described  :  also,  it  is  evident,  that  if  the  angle  ABD  be  greater  than  the 
angleBAD,  the  centre  E  falls  without  the  segment  ABC,  which  therefore 
is  less  than  a  semicircle  ;  but  if  the  angle  ABD  be  less  than  BAD,  the  cen- 
tre E  falls  within  the  segment  ABC,  which  is  therefore  greater  than  a  semi- 
circle :  Wherefore,  a  segment  of  a  circle  being  given,  the  circle  is  de- 
scribed of  which  it  is  a  segment. 

PROP.  XXVI.     THEOR. 

In  equal  circles,  equal  angles  stand  upon  equal  arcs,  whether  they  he  at  the 
centres  or  circumferences. 

Let  ABC,  DEF  be  equal  circles,  and  the  equal  angles  BGC,  EHF  at 
their  centres,  andBAC,  EDF  at  their  circumferences  :  the  arc  BKC  is 
equal  to  the  arc  ELF. 


78  ELEMENTS 

Join  BC,  EF  ;  and  because  the  circles  ABC,  DEF  are  equal,  the  straight 
lines  drawn  from  their  centres  are  equal  :  therefore  the  two  sides  BG 
GC,  are  equal  to  the  two  EH,  HF  ;  and  the  angle  at  G  is  equal  to  the  an- 
gle at  H  ;  therefore  the  base  BC  is  equal  (4.  1.)  to  the  base  EF  :  and  be- 
cause the  angle  at  A  is  equal  to  the  angle  at  D,  the  segment  BAC  is  similar 
(9.  def.  3.)  to  the  segment  EDF  ;  and  they  are  upon  equal  straight  lines 
BC,  EF  ;  but  similar  segments  of  circles  upon  equal  straight  lines  are 
equal  (24.  3.)  to  one  another,  therefore  the  segment  BAC  is  equal  to  the 
segment  EDF  :  but  the  whole  circle  ABC  is  equal  to  the  whole  DEF  ; 
therefore  the  remaining  segment  BKC  is  equal  to  the  remaining  segment 
ELF,  and  the  arc  BKC  to  the  arc  ELF. 


PROP.  XXVIL    THEOR. 

In  equal  circles,  the  angles  which  stand  upon  equal  arcs  are  equal  to  one 
another,  whether  they  he  at  the  centres  or  circumferences. 

Let  the  angles  BGC,  EHF  at  the  centres,  and  BAC,  EDF  at  the  cir- 
cumferences of  the  equal  circles  ABC,  DEF  stand  upon  the  equal  arcs 
BC,  EF  :  the  angle  BGC  is  equal  to  the  angle  EHF,  and  the  angle  BAC 
to  the  angle  EDF. 

If  the  angle  BGC  be  equal  to  the  angle  EHF,  it  is  manifest  (20.  3.) 
that  the  angle  BAC  is  also  equal  to  EDF.  But,  if  not,  one  of  them  is  the 
greater  :  let  BGC  be  the  greater,  and  at  the  point  G,  in  the  straight  line 
BG,  make  the  angle  (23.  L)  BGK  equal  to  the  angle  EHF.  And  because 
equal  angles  stand  upon  equal  arcs  (26.  3.),  when  they  are  at  the  centre. 


the  arc  BK  is  equal  to  the  arcjEF  :  but  EF  is  equalUo  BC  ;  therefore 
also  BK  is  equal  to  BC,  the  leSs  to  the  greater,  which  is  impossible.  There- 
fore the  angle  BGC  is  not  unequal  to  the  angle  EHF  ;  that  is,  it  is  equal 
to  it :  and  the  angle  at  A  is  half  the  angle  BGC,  and  the  angle  at  D  half 
of  the  angle  EHF  ;  therefore  the  angle  at  A  is  equal  to  the  angle  at  D. 

PROP.  XXVHL     THEOR. 

In  equal  circles,  equal  straight  lines  cut  off  equal  arcs,  the  greater  equal  to 
the  greater,  and  the  less  to  the  less. 

Let  ABC,  DEF  be  equal  circles,  and  BC,  EF  equal  straight  lines  in 
them,  which  cut  oflf  the  two  greater  arcs  BAC,  EDF,  and  the  two  less 


OF  GEOMETRY.     BOOK  III. 


79 


BGC,  EHF  :  the  greater  BAG  is  equal  to  the  greater  EDF,  and  the  less 
BGC  to  the  less  EHF. 

Take  (1.  3.)  K,  L,  the   centres  of  the  circles,  and  join  BK,  KG,  EL, 
LF  ;  and  because  the  circles  are  equal,  the  straight  lines  from  their  centres 


are  equal ;  therefore  BK,  KG  are  equal  to  EL,  LF ;  but  the  base  BC  is 
also  equal  to  the  base  EF  ;  therefore  the  angle  BKG  is  equal  (8.  1.)  to  the 
angle  ELF  :  and  equal  angles  stand  upon  equal  (26.  3.)  arcs,  when  they 
are  at  the  centres  ;  therefore  the  arc  BGG  is  equal  to  the  arc  EHF. 
But  the  whole  circle  ABG  is  equal  to  the  whole  EDF  ;  the  remaining  part, 
therefore,  of  the  circumference  viz.  BAG,  is  equal  to  the  remaining  part 
EDF. 


PROP.  XXIX.     THEOR. 

In  equal  circles  equal  arcs  are  subtended  by  equal  straight  lines. 

Let  ABG,  DEF  be  equal  circles,  and  let  the  arcs  BGC,  EHF  also  be 
equal ;  and  join  BG,  EF  :  the  straight  line  BG  is  equal  to  the  straight  line 
EF. 

Take  (1.3.)  K,  L  the  centres  of  the  circles,  and  join  BK,  KG,  EL,  LF  : 
and  because  the  arc  BGG  is  equal  to  the  arc  EHF,  the  angle  BKG  is 
equal  (27.  3.)  to  the  angle  ELF  :  also  because  the  circles  ABG,  DEF  are 
equal,  their  radii  are  equal :  therefore  BK,  KG  are  equal  to  EL,  LF  :  and 


C      E 


they  contain  equal  angles  ;  therefore  the  base  BG  is  equal  (4.  1.)  to  the 
base  EF. 


80 


ELEMENTS 


PROP.  XXX.     THEOR. 

To  bisect  a  given  arc,  that  is,  to  divide  it  into  two  equal  parts. 

Let  ADB  be  the  given  arc  ;  it  is  required  to  bisect  it. 

Join  AB,  and  bisect  (10.  1.)  it  in  C  ;  from  the  point  C  draw  CD  at  right 

angles  to  AB,  and  join  AD,  DB  :  the  arc  ADB  is  bisected  in  the  point  D. 
Because  AC  is  equal  to  CB,  and  CD  common  to  the  triangle  ACD, 

BCD,  the  two  sides  AC,  CD  are  equal  to  the  J) 

two  BC,  CD  ;  and  the  angle  ACD  is  equal  to 

the  angle  BCD,  because   each  of  them  is  a 

right  angle  :  therefore  the  base  AD   is  equal 

(4.  1.)  to  the  base  BD.     But  equal  straight 

lines  cut  off  equal  arcs,  (28.  3.)  the  greater      i^  C  JS 

equal  to  the  greater,  and  the  less  to  the  less  ;  and  AD,  DB  are  each  of 

them  less  than  a  semicircle,  because  DC  passes  through  the  centre  (Cor. 

1.  3.) ;  wherefore  the  arc  AD  is  equal  to  the  arc  DB  :  and  therefore  the 

given  arc  ADB  is  bisected  in  D. 


SCHOLIUM. 

By  the  same  construction,  each  of  the  halves  AD,  DB  may  be  divided 
into  two  equal  parts ;  and  thus,  by  successive  subdivisions,  a  given  arc 
may  be  divided  into  four,  eight,  sixteen,  &c.  equal  parts. 


PROP.  XXXL     THEOR. 

In  a  circle,  the  angle  in  a  semicircle  is  a  right  angle  ;  but  the  angle  in  a  seg- 
ment greater  than  a  semicircle  is  less  than  a  right  angle  ;  and  the  angle  in 
a  segment  less  than  a  semicircle  is  greater  than  a  right  angle. 

Let  ABCD  be  a  circle,  of  which  the  diameter  is  BC,  and  centre  E  ; 
draw  CA  dividing  the  circle  into  the  segments  ABC,  ADC,  and  join  BA, 
AD,  DC  ;  the  angle  in  the  semicircle  BAC  is  a  right  angle  ;  and  the  an- 
gle in  the  segment  ABC,  which  is  greater  than  a  semicircle,  is  less  than  a 
right  angle  ;  and  the  angle  in  the  segment  ADC,  which  is  less  than  a  semi- 
circle, is  greater  than  a  right  angle. 

Join  AE,  and  produce  BA  to  F  ;  and  because  BE  is  equal  to  EA,  the 
angle  EAB  is  equal  (5.  1.)  to  EBA :  also 
because  AE  is  equal  to  EC,  the  angle  EAC 
is  equal  to  EC  A  ;  wherefore  the  whole  an- 
gle BAC  is  equal  to  the  two  angles  ABC, 
ACB.  But  FAC,  the  exterior  angle  of  the 
triangle  ABC,  is  also  equal  (32.  1.)  to  the 
two  angles  ABC,  ACB  ;  therefore  the  an- 
gle BAC  is  equal  to  the  angle  FAC,  and 
each  of  them  is  therefore  a  right  angle  (7. 
def.  1 .) ;  wherefore  the  angle  BAC  in  a  semi- 
circle is  a  right  angle. 


OF  GEOMETRY.     BOOK  III. 


81 


And  because  the  two  angles  ABC,  BAG  of  the  triangle  ABG  are  to- 
gether less  (17. 1.)  than  two  right  angles,  and  Bx\G  is  a  right  angle,  ABG 
must  be  less  than  a  right  angle ;  and  therefore  the  angle  in  a  segment 
ABC,  greater  than 'a  semicircle,  is  less  than  a  right  angle. 

Also  because  ABCD  is  a  quadrilateral  figure  in  a  circle,  any  two  of  its 
opposite  angles  are  equal  (22.  3.)  to  two  right  angles  ;  therefore  theangles 
ABC,  ADC  are  equal  to  two  right  angles  ;  and  ABC  is  less  than  a  right 
angle  ;  wherefore  the  other  ADC  is  greater  than  a  right  angle. 

CoR.  From  this  it  is  manifest,  that  if  one  angle  of  a  triangle  be  equal  to 
the  other  two,  it  is  a  right  angle,  because  the  angle  adjacent  to  it  is  equal 
to  the  same  two  ;  and  when  the  adjacent  angles  are  equal,  they  are  right 
angles. 

PROP.  XXXII.     THEOR. 


f 


If  a  straight  line  touch  a  circle^  and  from  the  point  of  contact  a  straight 
line  be  drawn  cutting  the  circle,  the  angles  made  by  this  line  with  the  line 
which  touches  the  circle,  shall  be  equal  to  the  angles  in  the  alternate  seg- 
ments of  the  circle. 

Let  the  straight  line  EF  touch  the  circle  ABCD  in  B,  and  from  the 
point  B  let  the  straight  line  BD  be  drawn  cutting  the  circle :  the  angles 
which  BD  makes  with  the  touching  line  EF  shall  be  equal  to  the  angles 
in  the  alternate  segments  of  the  circle  :  that  is,  the  angle  FBD  is  equal  to 
the  angle  which  is  in  the  segment  DAB,  and  the  angle  DBE  to  the  angle 
in  the  segment  BCD.  '*~~ 

From  the  point  B  draw  (11.  1.)  BA  at  right  angles  to  EF,  and  take  any 
point  C  in  the  arc  BD,  and  join  AD,  DC,  GB ;  and  because  the  straight 
line  EF  touches  the  circle  ABCD  in  the  point  B,  and  BA  is  drawn  at  right 
angles  to  the  touching  line,  from  the  point  of  contact  B,  the  centre  of  the 
circle  is  (19.  3.)  in  BA  ;  therefore  the  an- 
gle ADB  in  a  semicircle,  is  a  right  an- 
gle (31.  3.),  and  consquently  the  other  two 
angles,  BAD,  ABD,  are  equal  (32,  1.)  to 
a  right  angle  ;  but  ABF  is  likewise  a  right 
angle  ;  therefore  the  angle  ABF  is  equal 
to  the  angles  BAD,  ABD  :  take  from 
these  equals  the  common  angle  ABD, 
and  there  will  remain  the  angle  DBF 
equal  to  the  angle  BAD,  which  is  in  the 
alternate  segment  of  the  circle.  And  be- 
cause ABCD  is  a  quadrilateral  figure  in 
a  circle,  the  opposite  angles  BAD,  BCD  are  equal  (22.  3.)  to  two  right 
angles  ;  therefore  the  angles  DBF,  DBE,  being  likewise  equal  (13.  1.)  to 
two  right  angles,  are  equal  to  the  angles  BAD,  BCD  ;  and  DBF  has  been 
proved  equal  to  BAD  :  therefore  the  remaining  angle  DBE  is  equal  to  the 
angle  BCD  in  the  alternate  segment  of  the  circle. 

11 


sa 


ELEMENTS 


PROP.  XXXIII.     PROB. 

Upon  a  given  straight  line  to  describe  a  segment  of  a  circle^  containing  hn 
angle  equal  to  a  givjBn  rectilineal  angle. 


Let  AB  be  the  given  straight  line,  and  the  angle  at  C  the  given  recti- 
lineal angle  ;  it  is  required  to  describe  upon  the  given  straight  line  AB  a 
segment  of  a  circle,  containing  an  angle  equal  to  the  angle  C. 

First,  let  the  angle  at  C  be  a  right  angle  ;  bisect  (10. 1.)  AB  in  F,  and 
from  the  centre  F,  at  the  distance  FB, 
describe  the  semicircle  AHB  ;  the  an- 
gle AHB  being  in  a  semicircle  is  (31. 
3.)  equal  to  the  right  angle  at  C. 

But  if  the  angle  C  be  not  a  right  an- 
gle at  the  point  A,  in  the  straight  line 
AB,  make  (23. 1 .)  the  angle  BAD  equal 

to  the  angle  C,  and  from  the  point  A  draw  (11.  1.)  AE  at  right  angles  to 
AD  ;  bisect  (10.  1.)  AB  in  F,  and 
from  F  draw  (11.  1.)  FG  at  right 
angles  to  AB,  and  join  GB  :  then 
because  AF  is  equal  to  FB,  and 
FG  common  to  the  triangles  AFG, 
BFG,  the  two  sides  AF,  FG  are 
equal  to  the  two  BF,  FG ;  but  the  ^ 
angle  AFG  is  also  equal  to  the  ^V 
angle  BFG ;  therefore  the  base  AG  ' 
is  equal  (4. 1.)  to  the  base  GB ;  and 
the  circle  described  from  the  centre 
G,  at  the  distance  GA,  shall  pass 
through  the  point  B  ;  let  this  be  the  circle  AHB :  and  because  from  the 
point  A  the  extremity  of  the  diameter  AE,  AD  is  drawn  at  right  angles  to 
AE,  therefore  AD  (Cor.  1.16. 3.)  touches 
the  circle  ;  and  because  AB,  drawn  from 
the  point  of  contact  A,  cuts  the  circle, 
the  angle  DAB  is  equal  to  the  angle  in 
the  alternate  segment  AHB  (32.  3.)  ; 
but  the  angle  DAB  is  equal  to  the  angle 
C,  therefore  also  the  angle  C  is  equal  to 
the  angle  in  the  segment  AHB  :  Where- 
fore, upon  the  given  straight  line  AB 
the  segment  AHB  of  a  circle  is  describ- 
ed which  contains  an  angle  equal  to  the  given  angle  at  C. 


OF  GEOMETRY.     BOOK  III. 


83 


PROP.  XXXIV.     PROB. 

To  cut  off  a  segment  from  a  given  circle  which  shall  contain  an  angle  equal 
to  a  given  rectilineal  angle. 

Let  ABC  be  the  given  circle,  and  D  the  given  rectilineal  angle  ;  it  is 
required  to  cut  off  a  segment  from  the  circle  ABC  that  shall  contain  an 
angle  equal  to  the  angle  D. 

Draw  (17.  3.)  the  straight  line  EF  touching  the  circle  ABC  in  the  point 
B,  and  at  the  point  B,  in  the  straight 
line  BF  make  (23. 1.)  the  angle  FBC 
equal  to  the  angle  D  ;  therefore,  be- 
cause the  straight  line  EF  touches 
the  circle  ABC,  and  BC  is  drawn 
from  the  point  of  contact  B,  the  an- 
gle FBC  is  equal  (32.  3.)  to  the  an- 
gle in  the  alternate  segment  BAC ; 
but  the  angle  FBC  is  equal  to  the  an- 
gle D  :  therefore  the  angle  in  the 
segment  BAC  is  equal  to  the  angle 
D  :  wherefore  the  segment  BAC  is  cut  off  from  the  given  circle  ABC 
containing  an  angle  equal  to  the  given  angle  D. 

PROP.  XXXV.     THEOR. 

If  two  straight  lines  within  a  circle  cut  one  another ^  the  rectangle  contained 
by  the  segments  of  one  of  them  is  equal  to  the  rectangle  contained  by  the 
segments  of  the  other, 

Le.t  the  two  straight  lines  AC,  BD,  within  the  circle  ABCD,  cut  one 
another  in  the  point  E  ;  the  rectangle  contained  by  AE,  EC  is  equal  to 
the  rectangle  contained  by  BE,  ED. 

If  AC,  BD  pass  each  of  them  through  the  cen- 
tre, so  that  E  is  the  centre,  it  is  evident  that  AE, 
EC,  BE,  ED,  being  all  equal,  the  rectangle  AE. 
EC  is  likewise  equal  to  the  rectangle  BE. ED. 

But  let  one  of  them  BD  pass  through  the  cen- 
tre, and  cut  the  other  AC,  which  does  not  pass 
through  the  centre,  at  right  angles  in  the  point  E  ; 
then,  if  BD  be  bisected  in  F,  F  is  the  centre  of 
the  circle  ABCD  ;  join  AF  :  and  because  BD,  which  passes  through  the 
centre,  cuts  the  straight  line  AC,  which  does  not 
pass  through  the  centre  at  right  angles,  in  E,  AE, 
EC  are  equal (3.  3.)  to  one  another;  and  because 
the  straight  line  BD  is  cut  into  two  equal  parts 
in  the  point  F,  and  into  two  unequal  in  the  point 
E,  BE.ED  (5.  2.)  +  EF2  =  FB'^  =  AF^.     But 
AF2  =  AE2  +  (47. 1.)  EF2,  therefore  BE.ED  + 
EF2,  ^  AE2  +  EF2,  and  taking  EF^  from  each, 
BE.ED=AE2=AE.EC. 

Next,  let  BD,  which  passes  through  the  centre, 
cut  the  other  AC,  which  does  not  pass  through 


84 


ELEMENTS 


the  centre,  in  E,  but  not  at  right  angles  ;  then,  as  before,  if  BD  be  bisect- 
ed in  F,  F  is  the  centre  of  the  circle.  Join  AF, 
and  from  F  draw  (12.  1.)  FG  perpendicular  to 
AC  ;  therefore  AG  is  equal  (3.  3.)  to  GC  ;  where- 
fore AE.EC  +  (5.  2.)  EG2  =  AG^,  and  adding 
GF2  to  both,  AE.EC  +  EG2+GF2=AG2+GF2. 
Now  EG2+GF2  =  EF2,  and  AG2+GF2=AF2 ; 
therefore  AE.EC  +  EF^^AF^zzzFB^.  But  FB2 
=BE.ED+(5.  2.)  EF2,  therefore  AE.EC+EF2 
=BE.ED+EF2,  and  taking  EF^  from  both,  AE. 
EC^BE.ED. 

Lastly,  let  neither  of  the  straight  lines  AC, 
BD  pass  through  the  centre :  take  the  centre  F, 
and  through  E,  the  intersection  of  the  straight 
lines  AC,  DB,  draw  the  diameter  GEFH :  and 
because,  as  has  been  shown,  AE.EC  =  GE.EH, 
and  BE.ED  =  GE.EH;  therefore  AE.EC=BE. 
ED. 


B     G 


PROP.  XXXVL     THEOR. 


If  from  any  point  without  a  circle  two  straight  lines  he  drawn,  one  of  which 
cuts  the  circle,  and  the  other  touches  it ;  the  rectangle  contained  hy  the  whole 
line  which  cuts  the  circle,  and  the  part  of  it  without  the  circle,  is  equal  to  the 
square  of  the  line  which  touches  it. 


Let  D  be  any  point  without  the  circle  ABC,  and  DCA,  DB  two  straight 
lines  drawn  from  it,  of  which  DCA  cuts  the  circle,  and  DB  touches  it : 
the  rectangle  AD. DC  is  equal  to  the  square  of  DB. 

Either  DCA  passes  through  the  centre,  or  it 
does  not ;  first,  let,  it  pass  through  the  centre  E, 
and  join  EB  ;  therefore  the  angle  EBD  is  a 
right  angle  (18.  3.) :  and  because  the  straight 
line  AC  is  bisected  in  E,  and  produced  to  the 
point  D,  AD.DC  +  EC2=ED2  (6.  2.).  But 
EC  =  EB,  therefore  AD.DC  +  EB2  =  ED2. 
Now  ED2=  (47. 1 .)  EB2-f  BD2,  because  EBD 
is  a  right  angle  ;  therefore  AD.DC  +  EB2  = 
EB2  4-BD2,  and  taking  EB2  from  each,  AD.DC 
=BD2. 

But,  if  DCA  does  not  pass  through  the  cen- 
tre of  the  circle  ABC,  take  (1.  3.)  the  centre  E, 
and  draw  EF  perpendicular  (12.  1.)  to  AC,  and 
join  EB,  EC,  ED  ;  and  because  the  straight 
line  EF,  which  passes  through  the  centre,  cuts 


OF  GEOMETRY.    BOOK  III. 


8$ 


the  straight  line  AC,  which  does  not  pass 
through  the  centre,  at  right  angles,  it  likewise 
bisects  it  (3.  3.) ;  therefore  AF  is  equal  to  FC  ; 
and  because  the  straight  line  AC  is  bisected  in 
F,  and  produced  to  D  (6.  2.),  AD.DC4-FC2= 
FD2 ;  add  FE2  to  both,  then  AD.DC  +  FC2+ 
FE2:=FD2-f-FE2.  But  (47.  1.)  EC2=FC2+ 
FE2,  and  ED2=FD24-FE2,  because  DFE  is 
aright  angle;  therefore  AD.DC  +  EC^^ED^. 
Now,  because  EBD  is  a  right  angle,  ED^i^ 
EB2-|-BD2=EC2+BD2,  and  therefore,  AD. 
DC-|-EC2=EC2-f  BD2,  and  AD.DCr=BD2. 

Cor.  1.  If  from  any  point  without  a  circle, 
there  be  drawn  two  straight  lines  cutting  it,  as 
AB,  AC,  the  rectangles  contained  by  the  whole 
lines  and  the  parts  of  them  without  the  circle, 
are  equal  to  one  another,  viz.  BA.AE  =  CA. 
AF  ;  for  each  of  these  rectangles  is  equal  to 
the  square  of  the  straight  line  AD,  which  touch- 
es the  circle. 

CoR.  2.  It  follows,  moreover,  that  two  tan- 
gents drawn  from  the  same  point  are  equal. 

Cor.  3.  And  since  a  radius  drawn  to  the 
point  of  contact  is  perpendicular  to  the  tangent, 
it  follows  that  the  angle  included  by  two  tangents^ 
drawn  from  the  same  point,  is  bisected  by  a  line 
drawn  from  the  centre  of  the  circle  to  that  point ; 
for  this  line  forms  the  hypotenuse  common  to 
two  equal  right  angled  triangles. 


PROP.  XXXVII.     THEOR. 


If  from  a  point  without  a  circle  there  be  drawn  two  straight  lines,  one  of 
which  cuts  the  circle,  and  the  other  meets  it ;  if  the  rectangle  contained  by 
the  whole  line,  which  cuts  the  circle,  and  the  part  of  it  without  the  circle, 
be  equal  to  the  square  of  the  line  which  meets  it,  the  line  which  meets  shall 
touch  the  circle. 

Let  any  point  D  be  taken  without  the  circle  ABC,  and  from  it  let  two 
straight  lines  DC  A  and  DB  be  drawn,  of  which  DC  A  cuts  the  circle,  and 
DB  meets  it ;  if  the  rectangle  AD.DC,  be  equal  to  the  square  of  DB,  DB 
touches  the  circle. 

Draw  (17.  3.)  the  straight  line  DE  touching  the  circle  ABC  ;  find  the 
centre  F,  and  join  FE,  FB,  FD  ;  then  FED  is  a  right  angle  (18.  3.) :  and 
because  DE  touches  the  circle  ABC,  and  DCA  cuts  it,  the  rectangle  AD, 
DC  is  equal  (36.  3.)  to  the  square  of  DE  ;  but  the  rectangle  AD.DC  is. 
by  hypothesis,  equal  to  the  square  of  DB  :  therefore  the  square  of  DE  is 


t 


86 


ELEMENTS 


equal  to  the  square  of  DB  ;  and  tlie  straight  line 
DE  equal  to  the  straight  line  DB  :  but  FE  is 
equal  to  FB,  wherefore  DE.EF  are  equal  to  DB, 
BF  ;  and  the  base  FD  is  common  to  the  two  trian- 
gles DEF,  DBF ;  therefore  the  angle  DEF  is 
equal  (8.  1.)  to  the  angle  DBF;  and  DEF  is  a 
right  angle,  therefore  also  DBF  is  a  right  angle  : 
but  FB,  if  produced,  is  a  diameter,  and  the  straight 
line  which  is  drawn  at  right  angles  to  a  diame- 
ter, from  the  extremity  of  it,  touches  (16.  3.)  the 
circle  :  therefore  DB  touches  the  circle  ABC. 


ADDITIONAL  PROPOSITIONS. 

PROP.  A.     THEOR. 

A  diameter  divides  a  circle  and  its  circumference  into  two  equal  parts;  and,con' 
versely,  the  line  which  divides  the  circle  into  two  equal   parts  is  a  diameter 

Let  AB  be  a  diameter  of  the  circle 
AEBD,  then  AEB,  ADB  are  equal  in 
surface  and  boundary. 

Now,  if  the  figure  AEB  be  applied  to 
the  figure  ADB,  their  common  base  AB 
retaining  its  position,  the  curve  line  AEB 
must  fall  on  the  curve  line  ADB  ;  other- 
wise there  would,  in  the  one  or  the  other, 
be  points  unequally  distant  from  the  cen- 
tre, which  is  contrary  to  the  definition  of 
a  circle. 


Conversely.     The  line  dividing  the  circle  into  two  equal  parts  is  a  diameter 

For,  let  AB  divide  the  circle  into  two  equal  parts  ;  then,  if  the  centre  is 
not  in  AB,  let  AF  be  drawn  through  it,  which  is  therefore  a  diameter,  and 
consequently  divides  the  circle  into  two  equal  parts  ;  hence  the  portion 
AEF  is  equal  to  the  portion  AEFB,  which  is  absurd. 

Cor,  The  arc  of  a  circle  whose  chord  is  a  diameter,  is  a  semicircum- 
ference,  and  the  included  segment  is  a  semicircle. 

PROP.  B.     THEOR. 

Through  three  given  points  which  are  not  in  the  same  straight  line,  one  cir- 
cumference of  a  circle  may  be  made  to  pass,  and  but  one. 

Let  A,  B,  C,  be  three  points  not  in  the  same  straight  line :  they  shall 
all  lie  in  the  same  circumference  of  a  circle. 


OF  GEOMETRY.    BOOK  III. 


87 


For,  let  the  distances  AB,  BC  be  bisected  by  the  perpendiculars  DF, 
EF,  which  must  meet  in  some  point  F  ;  for  if  they  were  parallel,  the  lines 
DB,  CB,  perpendicular  to  them  would  also  be  parallel  (Cor.  2.  29. 1.),  or 
else  form  but  one  straight  line :  but  they  meet  in  B,  and  ABC  is  not  a 
straight  line  by  hypothesis. 

Let  then,  FA,  FB,  and  FC  be  drawn ;  then, 
because  FA,  FB  meet  AB  at  equal  distances 
from  the  perpendicular,  they  are  equal.  For 
similar  reasons  FB,  FC,  are  equal ;  hence 
the  points  A,  B,  C,  are  all  equally  distant 
from  the  point  F,  and  consequently  lie  in  the 
circumference  of  the  circle,  whose  centre  is 
F,  and  radius  FA. 

It  is  obvious,  that  besides  this,  no  other 
circumference  can  pass  through  the  same 
points  ;  for  the  centre,  lying  in  the  perpen- 
'dicular  DF  bisecting  the  chord  AB,  and  at  the  same  time  in  the  perpen- 
dicular EF  bisecting  the  chord  BC  (Cor.  1.  3.  3.),  must  be  at  the  intersec- 
tion of  these  perpendiculars  ;  so  that,  as  there  is  but  one  centre,  there  can 
be  but  one  circumference. 

PROP.  C.    THEOR. 

If  two  circles  cut  each  other,  the  line  lohich  passes  through  their  centres  will  he 
perpendicular  to  the  chord  which  joins  the  points  of  intersection,  and  will 
divide  it  into  two  equal  parts. 

Let  CD  be  the  line  which  passes  through  the  centres  of  two  circles  cut- 
ting each  other,  it  will  be  perpendicular  to  the  chord  AB,  and  will  divide  it 
into  two  equal  parts. 

For  the  line  AB,  which  joins  the  points  of  intersection,  is  a  chord  com- 


mon to  the  two  circles.  And  if  a  perpendicular  be  erected  from  the  middle 
of  this  chord,  it  will  pass  (Cor.  1.  3.  3.)  through  each  of  the  two  centres  C 
and  D.  But  no  more  than  one  straight  line  can  be  drawn  through  two 
points  ;  hence,  the  straight  line  which  passes  through  the  centres  will  bi- 
sect the  chord  at  right  angles. 

CoR.     Hence,  the  line  joining  the  intersections  of  the  circumferences  of 
two  circles,  will  be  perpendicular  to  the  line  which  joins  their  centres. 

SCHOLIUM. 

1.  If  two  circles  cut  each  other,  the  distance  between  their  centres  will 
be  less  than  the  sum  of  their  radii,  and  the  greater  radius  will  be  also  less 


88 


ELEMENTS 


than  the  sum  of  the  smaller  and  the  distance  between  the  centres.  For, 
CD  is  less  (20.  1.)  than  CA-fAD,  and  for  the  same  reason,  AD/ AG  + 
CD. 

2.  And,  conversely,  if  the  distance  between  the  centres  of  two  circles 
be  less  than  the  sum  of  their  radii,  the  greater  radius  being  at  the  same  time 
less  than  the  sum  of  the  smaller  and  the  distance  between  the  centres, 
the  two  circles  will  cut  each  other. 

For,  to  make  an  intersection  possible,  the  triangle  CAD  must  be  possi- 
ble. Hence,  not  only  must  we  have  CD<AC-}-AD,  but  also  the  greater 
radius  AD<;AC-{-CI)  ;  And  whenever  the  triangle  CAD  can  be  con- 
structed, it  is  plain  that  the  circles  described  from  the  centres  C  and  D, 
will  cut  each  other  in  A  and  B. 

Cor.  1.  Hence,  if  the  distance  between  the  centres  of  two  circles  be 
greater  than  the  sum  of  their  radii,  the  two  circles  will  not  intersect  each 
other. 

Cor.  2.  Hence,  also,  if  the  distance  between  the  centres  be  less  than 
the  difference  of  the  radii,  the  two  circles  will  not  cut  each  other. ' 

For,  AC4-CD>AD;  therefore,  CD>AD-AC  ;  that  is,  any  side  of 
a  triangle  exceeds  the  difference  between  the  other  two.  Hence,  the  tri- 
angle is  impossible  when  the  distance  between  the  centres  is  less  than  the 
difference  of  the  radii ;  and  consequently  the  two  circles  cannot  cut  each 
other. 

PROP.  D.    THEOR. 

In  the  same  circle^  equal  angles  at  the  centre  are  subtended  hy  equal  arcs  ; 
andf  conversely,  equal  arcs  subtend  equal  angles  at  the  centre. 

Let  C  be  the  centre  of  a  circle,  and  let  the  angle  ACD  be  equal  to  the 
angle  BCD  ;  then  the  arcs  AFD,  DGB,  subtending  these  angles,  are 
equal. 

Join  AD,  DB  ;  then  the  triangles  ACD, 
BCD,  having  two  sides  and  the  included  an- 
gle in  the  one,  equal  to  two  sides  and  the 
included  angle  in  the  other,  are  equal :  so 
that,  if  ACD  be  applied  to  BCD,  there  shall 
be  an  entire  coincidence,  the  point  A  coin- 
ciding with  B,  and  D  common  to  both  arcs  ; 
the  two  extremities,  therefore,  of  the  arc 
AFD,  thus  coinciding  with  those  of  the  arc 
BGD,  all  the  intermediate  parts  must  coin- 
cide, inasmuch  as  they  are  all  equally  dis- 
tant from  the  centre. 

Conversely.  Let  the  arc  AFD  be  equal  to  the  arc  BGD  ;  then  the  an- 
gle ACD  is  equal  to  the  angle  BCD. 

For,  if  the  arc  AFD  be  applied  to  the  arc  BGD,  they  would  coincide  ; 
so  that  the  extremities  AD  of  the  chord  AD,  would  coincide  with  those  of 
the  chord  BD  ;  these  chords  are  therefore  equal  :  hence,  the  angle  ACD 
is  equal  to  the  angle  BCD  (8.  1.). 

CoR.  L     It  follows,  moreover,  that  equal  angles  at  the  centre  are  sub- 


OF  GEOMETRY.     BOOK  III.  89 

tended  by  equal  chords  :  and,  conversely,  equal  chords  subtend  equal  an- 
gles at  the  centre. 

Cor.  2.  It  is  also  evident,  that  equal  chords  subtend  equal  arcs  :  and, 
conversely,  equal  arcs  are  subtended  by  equal  chords. 

Cor.  3.  If  the  angle  at  the  centre  of  a  circle  be  bisected,  both  the  arc 
and  the  chord  which  it  subtends  shall  also  be  bisected. 

Cor.  4.  It  follows,  likewise,  that  a  perpendicular  through  the  middle 
of  the  chord,  bisects  the  angle  at  the  centre,  and  passes  through  the  middle 
of  the  arc  subtended  by  that  chord. 

SCHOLIUM. 

The  centre  C,  the  middle  point  E  of  the  chord  AB,  and  the  middle  point 
D  of  the  arc  subtended  by  this  chord,  are  three  points  situated  in  the  same 
line  perpendicular  to  the  chord.  But  two  points  are  sufficient  to  determine 
the  position  of  a  straight  line  ;  hence  every  strtiight  line  which  passes 
through  two  of  the  points  just  mentioned,  will  necessarily  pass  through  the 
third,  and  be  perpendicular  to  the  chord. 

PROP.  E.    THEOR. 

The  arcs  of  a  circle  intercepted  by  two  parallels  are  equal ;  and^  conversely,  if 
two  straight  lines  intercept  equal  arcs  of  a  circle^  and  do  not  cut  each  other 
within  the  circle^  the  lines  will  be  parallel. 

There  may  be  three  cases  : 

First.  If  the  parallels  are  tangents 
to  the  circle,  as  AB,  CD  ;  then,  each 
of  the  arcs  intercepted  is  a  semi-cir- 
cumference, as  their  points  of  contact 
(Cor,  3.  16.  3.)  coincide  with  the  ex- 
tremities of  the  diameter. 

Second.  When,  of  the  two  parallels 
AB,  GH,  one  is  a  tangent,  the  other 
a  chord,  which  being  perpendicular  to 
FE,  the  arc  GEH  is  bisected  by  FE 
(Cor.  4.  Prop.  D.  Book  3.) ;  so  that  in 
this  case  also,  the  intercepted  arcs 
GE,  EH  are  equal. 

Third.  If  the  two  parallels  are  chords,  as  GH,  JK ;  let  the  diameter 
FE  be  perpendicular  to  the  chord  GH,  it  will  also  be  perpendicular  to  JK, 
since  they  are  parallel ;  therefore,  this  diameter  must  bisect  each  of  the 
arcs  which  they  subtend:  that  is,  GE  =  EH,  and  JE  =  EK  ;  therefore, 
JE  — GEzriEK— EH ;  or,  which  amounts  to  the  same  thing,  JG  is  equal 
to  HK.  ^  ^ 

Conversely.  If  the  two  lines  be  AB,  CD,  which  touch  the  circumfer- 
ence, and  if,  at  the  same  time,  the  intercepted  arcs  EJF,  EKF  are  equal, 
EF  must  be  a  diameter  (Prop.  A.  Book  3.) ;  and  therefore  AB,  CD  (Cor. 
3.  16.  3.),  are  parallel. 

But  if  only  one  of  the  lines,  as  AB,  touch,  while  the  other,  GH,  cuts  the 
circumference,  making  the  arcs  EG,  EH  equal;  then  the  diameter  FE, 

12 


90 


ELEMENTS,  &c. 


which  bisects  the  arc  GEH,  is  perpendicular  (Schol.  D.  3.)  to  its  chord 
GH  :  it  is  also  perpendicular  to  the  tangent  AB  ;  therefore  AB,  GH  are 
parallel. 

If  both  lines  cut  the  circle,  as  GH,  JK,  and  intercept  equal  arcs  GJ, 
HK ;  let  the  diameter  FE  bisect  one  of  the  chords,  as  GH  :  it  will  also 
bisect  the  arc  GEH,  so  that  EG  is  equal  to  EH  ;  and  since  GJ  is  {by  hyp.) 
equal  to  HK,  the  whole  arc  EJ  is  equal  to  the  whole  arc  EK  ;  therefore 
the  chord  JK  is  bisected  by  the  diameter  FE  :  hence,  as  both  chords  are 
bisected  by  the  diameter  FE,  they  are  perpendicular  to  it ;  that  is,  they  are 
parallel  (Cor.  28  1.). 

SCHOLIUM. 

The  restriction  in  the  enunciation  of  the  converse  proposition,  namely, 
that  the  lines  do  not  cut  each  other  within  the  circle,  is  necessary ;  for 
lines  drawn  through  the  points  G,  K,  and  J,  H,  will  intercept  equal  arcs 
GJ,  HK,  and  yet  not  be  parallel,  since  they  will  intersect  each  other  within 
the  circle. 

PROP.  F.     PROB. 
To  draw  a  tangent  to  any  point  in  a  circular  arCf  without  Jinding  the  centre. 

From  B  the  given  point,  take  two  equal 
distances  BC,  CD  on  the  arc  ;  join  BD, 
and  draw  the  chords  BC,  CD  :  make  (23. 
1.)  the  angle  CBG=CBD,  andthe  straight 
line  BG  will  be  the  tangent  required. 

For  the  angle  CBD=:CDB  ;  and  there- 
fore the  angle  GBC  (32.  3.)  is  also  equal 
to  CD B,  an  angle  in  .he  alternate  segment ; 
hence,  BG  is  a  tangent  at  B. 


ELEMENTS 


OF 


GEOMETRY. 


BOOK  IV. 


liW"  DEFINITIONS.  ' 

1  A  RECTILINEAL  figure  is  said  to  be  inscribed  in  another  rectilineal 
figure,  when  all  the  angles  of  the  inscribed 

figure  are  upon  the  sides  of  the  figure  in  which 
it  is  inscribed,  each  upon  each. 

2  In-  like  manner,  a  figure  is  said  to  be  described 
about  another  figure,  when  all  the  sides  of  the 
circumscribed  figure  pass  through  the  angular 
points  of  the  figure  about  which  it  is  described, 
each  through  each. 

3  A  rectilineal  figure  is  said  to  be  inscribed  in 
a  circle,  when  all  the  angles  of  the  inscribed 
figure  are  upon  the  circumference  of  the  cir- 
cle. 

4.  A  rectilineal  figure  is  said  to  be  described 
about  a  circle,  when  each  side  of  the  circum- 
scribed figure  touches  the  circumference  of  the 
circle. 

5.  In  like  manner,  a  circle  is  said  to  be  inscrib- 
ed in  a  rectilineal  figure,  when  the  circum- 
ference of  the  circle  touches  each  side  of  the 
figure. 

6.  A  circle  is  said  to  be  described  about  a  recti- 
lineal figure,  when  the  circumference  of  the 
circle  passes  through  all  the  angular  points  of 
the  figure  about  which  it  is  described. 

7.  A  straight  line  is  said  to  be  placed  in  a  circle, 
when  the  extremities  of  it  are  in  the  circum- 
ference of  the  circle. 


92  ELEMENTS 

8.  Polygons  of  five  sides  are  called  pentagons ;  those  of  six  sides,  hexc' 
gons ;  those  of  seven  sides,  heptagons ;  those  of  eight  sides,  octagons  / 
and  so  on. 

9.  A  polygon,  which  is  at  once  equilateral  and  equiangular,  is  called  a 
regular  polygon. 

Regular  polygons  may  have  any  number  of  sides ;  the  equilateral  tri 
angle  is  one  of  three  sides ;  and  the  square  is  one  of  four  sides. 

LEMMA. 

Any  regular  polygon  may  he  inscribed  in  a  circle,  and  circumscribed  about  one. 

Let  ABCDE,  &c.  be  a  regular  polygon :  describe  a  circle  through  the 
three  points  A,  B,  C,  the  centre  being  0,  and  OP  the  perpendicular  let  fall 
from  it,  to  the  middle  point  of  BC  :  join  AO  and  OD. 

If  the  quadrilateral  OPCD  be  placed  upon 
the  quadrilateral  OPBA,  they  will  coincide ; 
for  the  side  OP  is  common :  the  angle  OPC=s 
OPB,  being  right ;  hence  the  side  PC  will  ap- 
ply to  its  equal  PB,  and  the  point  C  will  fall 
on  B  ;  besides,  from  the  nature  of  the  polygon, 
the  angle  PCD=PBA;  hence  CD  will  take 
the  direction  BA,  and  since  CD=BA,the  point 
D  will  fall  on  A,  and  the  two  quadrilaterals 
will  entirely  coincide.  ^^ 

The  distance  OD  is  therefore  equal  to  AO  ;  a* 

and  consequently  the  circle  which  passes  through  the  three  points  A,  B,  C, 
will  also  pass  through  the  point  D.  By  the  same  mode  of  reasoning,  it 
might  be  shown  that  the  circle  which  passes  through  the  points  B,  C,  D, 
will  also  pass  through  the  point  E  ;  and  so  of  all  the  rest:  hence  the  cir- 
cle which  passes  through  the  points  A,  B,  C,  passes  through  the  vertices 
of  all  the  angles  in  the  polygon,  which  is  therefore  inscribed  in  this  circle. 

Again,  in  reference  to  this  circle,  all  the  sides  AB,  BC,  CD,  &:c.  are 
equal  chords  ;  they  are  therefore  equally  distant  from  the  centre  (Th.  14. 
3.) :  hence,  if  from  the  point  O  with  the  distance  OP,  a  circle  be  describ- 
ed, it  will  touch  the  side  BC,  and  all  the  other  sides  of  the  polygon,  each 
in  its  middle  point,  and  the  circle  will  be  inscribed  in  the  polygon,  or  the 
polygon  circumscribed  about  the  circle. 

CoR.  L  Hence  it  is  evident  that  a  circle  may  be  inscribed  in,  or  cir- 
cumscribed about,  any  regular  polygon,  and  the  circles  so  described  have  a 
common  centre.  \ 

CoR.  2.  Hence  it  likewise  follows,  that  if  from  a  common  centre,  circles 
can  be  inscribed  in,  and  circumscribed  about  a  polygon,  that  polygon  is  regu- 
lar. For,  supposing  those  circles  to  be  described,  the  inner  one  will  touch 
all  the  sides  of  the  polygon  ;  these  sides  are  therefore  equally  distant  from 
its  centre  ;  and,  consequently,  being  chords  of  the  circumscribed  circle, 
they  are  equal,  and  therefore  include  equal  angles.  Hence  the  polygon  is 
at  once  equilateral  and  equiangular ;  that  is  (Def.  9.  B.  IV.),  it  is  regular. 


OF  GEOMETRY.    BOOK  IV.  93 

SCHOLIUMS. 

1.  The  point  0,  the  common  centre  of  the  inscribed  and  circumscribed 
circles,  may  also  be  regarded  as  the  centre  of  the  polygon  ;  and  upon  this 
principle  the  angle  AOB  is  called  the  angle  at  the  centre^  being  formed  by 
two  radii  drawn  to  the  extremities  of  the  same  side  AB. 

Since  all  the  chords  are  equal,  all  the  angles  at  the  centre  must  evident- 
ly be  equal  likewise  ;  and  therefore  the  value  of  each  will  be  found  by  di- 
viding four  right  angles  by  the  number  of  the  polygon's  sides. 

2.  To  inscribe  a  regular  polygon  of  a  certain  number  of  sides  in  a  given 
circle,  we  have  only  to  divide  the  circumference  into  as  many  equal  parts 
as  the  polygon  has  sides  :  for  the  arcs  being  equal  (see  fig.  Prop.  XV.  B.  4.), 
the  chords  AB,  BO,  CD,  &;c.  will  also  be  equal ;  hence,  likewise,  the  tri- 
angles ABG,  BGC,  CGD,  &c.  must  be  equal,  because  they  are  equian- 
gular ;  hence  all  the  angles  ABC,  BCD,  CDE,  &c.  will  be  equal,  and  con- 
sequently the  figure  ABCD,  &c.  will  be  a  regular  polygon. 

PROP.  I.    PROB. 

In  a  given  circle  to  place  a  straight  line  equal  to  a  given  straight  line^  not 
greater  than  the  diameter  of  the  circle. 

Let  ABC  be  the  given  circle,  and  D  the  given  straight  line,  not  greater 
than  the  diameter  of  the  circle. 

Draw  BC  the  diameter  of  the  circle 
ABC  :  then,  if  BC  is  equal  to  D,  the 
thing  required  is  done  ;  for  in  the  circle 
ABC  a  straight  line  BC  is  placed  equal 
to  D  ;  But,  if  it  is  not,  BC  is  greater 
than  D  ;  make  CE  equal  (Prop.  3.  I.) 
to  D,  and  from  the  centre  C,  at  the  dis- 
tance CE,  describe  the  circle  AEF,  and 
join  CA :  Therefore,  because  C  is  the 
centre  of  the  circle  AEF,  CA  is  equal 
to  CF  ;  but  D  is  equal  to  CE  ;  there- 
fore D  is  equal  to  CA  :  Wherefore,  in  the  circle  ABC,  a  straight  line  is 
placed,  equal  to  the  given  straight  line  D,  which  is  not  greater  than  the 
diameter  of  the  circle. 


PROP.  II.     PROB. 
In  a  given  circle  to  inscribe  a  triangle  equiangular  to  a  given  tnangle. 

Let  ABC  be  the  given  circle,  and  DEF  the  given  triangle  ;  it  is  re- 
quired to  inscribe  in  the  circle  ABC  a  triangle  equiangular  to  the  triangle 
DEF. 

Draw  (Prop.  17.  3.)  the  straight  line  GAH  touching  the  circle  in  the  point 
A,  and  at  the  point  A,  in  the  straight  line  AH,  make  (Prop.  23.  l.)the  an- 
gle HAC  equal  to  the  angle  DEF  ;  and  at  the  point  A,  in  the  straight  line 


94 


ELEMENTS 


AG,  make  the  angle  GAB  equal 
to  the  angle  DFE,  and  join 
BC.  Therefore,  because  HAG* 
touches  the  circle  ABC,  and  AC 
is  drawn  from  the  point  of  con- 
tact, the  angle  HAC  is  equal 
(32.  3.)  to  the  angle  ABC  in  the 
alternate  segment  of  the  circle  : 
But  HAC  is  equal  to  the  angle 
DEF  ;  therefore  also  the  angle 
ABC  is  equal  to  DEF  ;  for  the 
same  reason,  the  angle  ACB  is 
equal  to  the  angle  DFE  ;  therefore  the  remaining  angle  BAG  is  equal 
(4.  Cor.  32.  1.)  to  the  remaining  angle  EDF :  Wherefore  the  triangle  ABC 
is  equiangular  to  the  triangle  DEF,  and  it  is  inscribed  in  the  circle  ABG 


PROP.  HI.     PROB. 
About  a  given  circle  to  describe  a  triangle  equiangular  to  a  given  triangle. 

Let  ABC  be  the  given  circle  and  DEF  the  given  triangle  ;  it  is  requir- 
ed to  describe  a  triangle  about  the  circle  ABC  equiangular  to  the  triangle 
DEF. 

Produce  EF  both  ways  to  the  points  G,  H,  and  find  the  centre  K  of  the 
circle  ABC,  and  from  it  draw  any  straight  line  KB ;  at  the  point  K  in  the 
straight  line  KB,  make  (Prop.  23  1.)  the  angle  BKA  equal  to  the  angle 
DEG,  and  the  angle  BKC  equal  to  the  angle  DFH  ;  and  through  the 
points  A,  B,  C,  draw  the  straight  lines  LAM,  MBN,  NCL  touching  (Prop. 
17.3.)  the  circle  ABC  :  Therefore,  because  LM,  MN,  NL  touch  the  circle 
ABC  in  the  points  A,  B,  C,  to  which  from  the  centre  are  drawn  KA,  KB, 
KG,  the  angles  at  the  points  A,  B,  C,  are  right  (18.  3.)  angles.  And  be- 
cause the  four  angles  of  the  quadrilateral  figure  AMBK  are  equal  to  four 
right  angles,  for  it  can  be  divided  into  two  triangles  ;  and  because  two  of 


them,  KAM,  KBM,  are  right  angles,  the  other  two  AKB,  AMB  are  equal 
to  two  right  angles  ;  But  the  angles  DEG,  DEF  are  likewise  equal  (13.1.) 
to  two  right  angles  ;  therefore  the  angles  AKB,  AMB  are  equal  to  the  an- 
gles DEG,  DEF,  of  which  AKB  is  equal  to  DEG  ;  wherefore  the  remain- 


OF  GEOMETRY.     BOOK  IV. 


95 


ing  angle  AMB  is  equal  to  the  remaining  angle  DEF.  In  like  manner, 
the  angle  LMN  may  be  demonstrated  to  be  equal  to  DFE  ;  and  therefore 
the  remaining  angle  MLN  is  equal  (32.  1.)  to  the  remaining  angle  EDF  : 
Wherefore  the  triangle  LMN  is  equiangular  to  the  triangle  DEF  :  and  it 
is  described  about  the  circle  ABC. 

PROP.  IV.     PROB. 

To  inscribe  a  circle  in  a  given  triangle. 

Let  the  given  triangle  be  ABC  ;  it  is  required  to  inscribe  a  circle  in 
ABC. 

Bisect  (9.  1.)  the  angles  ABC,  BCA  by  the  straight  lines  BD,  CD  meet- 
ing one  another  in  the  point  D,  from  which  draw  (12.  1.)  DE,  DF,  DG 
perpendiculars  to  AB,  BC,  CA.  Then  be- 
cause the  angle  EBD  is  equal  to  the  angle 
FBD,  the  angle  ABC  being  bisected  by 
BD  ;  and  because  the  right  angle  BED,  is 
equal  to  the  right  angle  BFD,  the  two  tri- 
angles EBD,  FBD  have  two  angles  of  the 
one  equal  to  two  angles  of  the  other  ;  and 
the  side  BD,  which  is  opposite  to  one  of 
the  equal  angles  in  each,  is  common  to 
both ;  therefore  their  other  sides  are  equal 
(26.  1.);  wherefore  DE  is  equal  to  DF. 
For  the  same  reason,  DG  is  equal  to 
DF ,  therefore  the  three  straight  lines  DE,  DF,  DG,  are  equal  to  one 
another,  and  the  circle  described  from  the  centre  I),  at  the  distance  of  any 
of  them,  will  pass  through  the  extremities  of  the  other  two,  and  will  touch 
the  straight  lines  AB,  BC,  CA,  because  the  angles  at  the  points  E,  F,  G, 
are  right  angles,  and  the  straight  line  which  is  drawn  from  the  extremity 
of  a  diameter  at  right  angles  to  it,  touches  (1  Cor.  16. 3.)  the  circle.  There- 
fore the  straight  lines  AB,  BC,  CA,  do  each  of  them  touch  the  circle,  and 
the  circle  EFG  is  inscribed  in  the  triangle  ABC. 

PROP.  V.     PROB. 

To  describe  a  circle  about  a  given  triangle.  , 

Let  the  given  triangle  be  ABC  ;  it  is  required  to  describe  a  circle  about 
ABC. 

Bisect  flO.  1.)  AB,  AC  in  the  points  D,  E,  and  from  these  points  draw 


96  ELEMENTS 

DF,  EF  at  right  angles  (11.  1.)  to  AB,  AC  ;  DF,  EF  produced  will  meet 
one  another;  for,  if  they  do  not  meet,  they  are  parallel,  wherefore,  AB, 
AC,  which  are  at  right  angles  to  them,  are  parallel,  which  is  absurd  :  let 
them  meet  in  F,  and  join  FA  ;  also,  if  the  point  F  be  not  in  BC,  join  BF, 
CF  :  then,  because  AD  is  equal  to  BD,  and  DF  common,  and  at  right  an- 
gles to  AB,  the  base  AF  is  equal  (4.  1.)  to  the  base  FB.  In  like  manner, 
it  may  be  shewn  that  CF  is  equal  to  FA;  and  therefore  BF  is  equal  to 
FC  ;  and  FA,  FB,  FC  are  equal  to  one  another  ;  wherefore  the  circle  de- 
scribed from  the  centre  F,  at  the  distance  of  one  of  them,  will  pass 
through  the  extremities  of  the  other  two,  and  be  described  about  the  trian- 
gle ABC. 

Cor.  When  the  centre  of  the  circle  falls  within  the  triangle,  each  of 
its  angles  is  less  than  a  right  angle,  each  of  them  being  in  a  segment  great- 
er than  a  semicircle  ;  but  when  the  centre  is  in  one  of  the  sides  of  the 
triangle,  the  angle  opposite  to  this  side,  bfing  in  a  semicircle,  is  a  right  an- 
gle :  and  if  the  centre  falls  without  the  triangle,  the  angle  opposite  to  the 
side  beyond  which  it  is,  being  in  a  segment  less  than  a  semicircle,  is  greater 
than  a  right  angle.  Wherefore,  if  the  given  triangle  be  acute  angled,  the 
centre  of  the  circle  falls  within  it ;  if  it  be  a  right  angle  triangle,  the  cen- 
tre is  in  the  side  opposite  to  the  right  angle  ;  and  if  it  be  an  obtuse  angled 
triangle,  the  centre  falls  without  the  triangle,  beyond  the  side  opposite  to  the 
obtuse  angle. 

SCHOLIUM. 

1.  From  the  demonstration  it  is  evident  that  the  three  perpendiculars 
bisecting  the  sides  of  a  triangle,  meet  in  the  same  point ;  that  is,  the  centre 
of  the  circumscribed  circle. 

2.  A  circular  segment  arch  of  a  given  span  and  rise,  may  be  drawn  by 
a  modification  of  the  preceding  problem. 

Let  AB  be  the  span  and  SR  the  rise. 

Join  AR,  BR,  and  at  their  respective  points  of  bisection,  M,  N,  erect 
the  perpendicular  MO,  NO  to  AR,  BR  ;  they 
will  intersect  at  O,  the  centre  of  the  circle. 
That  OA  =  OR  =  OB,  is  proved  as  before. 

The  joints  betweert  the  arch-stones,  or 
voussoirs,  are  only  continuations  of  radii 
drawn  from  the  centre  0  of  the  circle. 


PROP.  VI.     PROB. 
To  inscribe  a  square  in  a  given  circle. 

Let  ABCD  be  the  given  circle ;  it  is  required  to  inscribe  a  square  in 
ABCD. 

Draw  the  diameters,  AC,  BD  at  right  angles  to  one  another,  and  join 
AB,  BC,  CD,  DA ;  because  BE  is  equal  to  ED,  E  being  the  centre,  and 


OF  GEOMETRY.     BOOK  IV. 


07 


because  EA.  is  at  right  angles  to  BD,  and 
common  to  the  triangles  ABE,  ADE  ;  the 
base  BA  is  equal  (4.  1.)  to  the  base  AD  ;  and, 
for  the  same  reason,  BC,  CD  are  each  of 
them  equal  to  BA  or  AD  ;  therefore  the  quad- 
rilateral figure  ABCD  is  equilateral.  It  is 
also  rectangular ;  for  the  straight  line  BD  be- 
ing a  diameter  of  the  circle  ABCD,  BAD  is 
a  semicircle  ;  wherefore  the  angle  BAD  is  a 
right  angle  (31.3.);  for  the  same  reason  each 
of  the  angles  ABC,  BCD,  CDA  is  a  right  an- 
gle ;  therefore  the  quadrilateral  figure  ABCD 
is  rectangular,  and  it  has  been  shewn  to  be 
equilateral ;  therefore  it  is  a  square  ;  and  it 
ABCD. 

SCHOLIUM. 


is  inscribed  in  the  circle 


Since  the  triangle  AED  is  right  angled  and  isosceles,  we  have  (Cor.  2. 
47.  1)  AD  :  AE  :  :  y2  :  1  ;  hence  the  side  of  the  inscribed  square  is  to 
the  radius,  as  the  square  root  of  2,  is  to  unity. 

PROP.  VII.     PROB. 

To  describe  a  square  about  a  given  circle. 

Let  ABCD  be  the  given  circle ;  it  is  required  to  describe  a  square  about  it. 

Draw  two  diameters  AC,  BD  of  the  circle  ABCD,  at  right  angles  to 
one  another,  and  through  the  points  A,  B,  C,  D  draw  (17.  3.)  FG,  GH,  HK, 
KF  touching  the  circle  ;  and  because  FG  touches  the  circle  ABCD,  and 
EA  is  drawn  from  the  centre  E  to  the  point  of  contact  A,  the  angles  at  A 
are  right  angles  (18.  3.) ;  for  the  same  reason,  the  angles  at  the  points  B, 
C,  D,  are  right  angles ;  and  because  the  angle  AEB  is  a  right  angle,  as 
likewise  i^  EBG,  GH  is  parallel  (28.  1.)  to  AC  ;  for  the  same  reason,  AC. 
is  parallel  to  FK,  and  in  like  manner,  GF, 
HK  may  each  of  them  be  demonstrated  to  be 
parallel  to  BED ;  therefore  the  figures  GK, 
GC,  AK,  FB,  BK  are  parallelograms ;  and 
GF  is  therefore  equal  (34.  1.)  to  HK,  and  GH 
to  FK ;  and  because  AC  is  equal  to  BD, 
and  also  to  each  of  the  two  GH,  FK ;  and 
BD  to  each  of  the  two  GF,  HK :  GH,  FK 
are  each  of  them  equal  to  GF  or  HK  ;  there- 
fore the  quadrilateral  figure  FGHK  is  equi- 
lateral. It  is  also  rectangular ;  for  GBEA 
being  a  parallelogram,  and  AEB  a  right  an- 
gle, AGB  (34.  1.)  is  likewise  a  right  angle  : 
in  the  same  manner,  it  may  be  shewn  that  the  angles  at  H,  K,  F  are  right 
angles ;  therefore  the  quadrilateral  figure  FGHK  is  rectangular  ;  and  it 
was  demonstrated  to  be  equilateral ;  therefore  it  is  a  square  ;  and  it  is  de- 
scribed about  the  circle  ABCD. 

13 


98 


ELEMENTS 


*  PROP.  VIII.     PROB. 

To  inscribe  a  circle  in  a  given  square. 

Let  ABCD  be  the  given  square  ;  it  is  required  to  inscribe  a  circle  in 
ABCD. 

Bisect  (10.  1.)  each  of  the  sides  AB,  AD,  in  the  points  F,  E,  and 
through  E  draw  (31.  1.)  EH  parallel  to  AB  or  DC,  and  through  F  draw 
FK  parallel  to  AD  or  BC  ;  therefore  each  of  the  figures,  AK,  KB,  AH, 
HD,  AG,  GC,  BG,  GD  is  a  parallelogram,  and  their  opposite  sides  are 
equal  (34.  1.) ;  and  because  that  AD  is  equal  to  AB,  and  that  AE  is  the 
half  of  AD,  and  AF  the  half  of  AB,  AE  is  equal  to  AF ;  wherefore  thie 
sides  opposite  to  these  are  equal,  viz.  FG  to  GE  ;  in  the  same  manner  it 
may  be  demonstrated,  that  GH,  GK,  are  each 
of  them  equal  to  FG  or  GE  ;  therefore  the 
four  straight  lines,  GE,  GF,  GH,  GK,  are 
equal  to  one  another ;  and  the  circle  described 
from  the  centre  G,  at  the  distance  of  one  of 
them,  will  pass  through  the  extremities  of  the 
other  three  ;  and  will  also  touch  the  straight 
lines  AB,  BC,  CD,  DA,  because  the  angles 
at  the  points  E,  F,  H,  K,  are  right  angles 
(29.  1.),  and  because  the  straight  line  which 
is  drawn  from  the  extremity  of  a  diameter  at 
right  angles  to  it,  touches  the  circle  (16.  3.) ; 
therefore  each  of  the  straight  lines  AB,  BC, 
CD,  DA  touches  the  circle,  which  is  therefore  inscribed  in  the  squares 
ABCD. 

PROP.  IX.     PROB. 
To  describe  a  circle  about  a  given  square. 

Let  ABCD  be  the  given  square  ;  it  is  required  to  describe  a  circle 
about  it. 

Join  AC,  BD,  cutting  one  another  in  E  ;  and  because  DA  is  equal  to 
AB,  and  AC  common  to  the  triangles  DAC,  BAC,  the  two  sides  DA,  AC 
are  equal  to  the  two  BA,  AC,  and  the  base  DC  is  equal  to  the  base  BC  ; 
wherefore  the  angle  DAC  is  equal  (8.  1.)  to  the 
angle  BAC,  and  the  angle  DAB  is  bisected  by 
the  straight  line  AC.    In  the  same  manner  it  may 
be  demonstrated,  that  the   angles  ABC,  BCD, 
CD  A  are  severally  bisected  by  the  straight  lines 
BD,  AC  ;  therefore,  because  the  angle  DAB  is 
equal  to  the  angle  ABC,  and  the  angle  EAB  is 
the  half  of  DAB,  and  EB A  the  half  of  ABC  ;  the 
angle  EAB  is  equal  to  the  angle  EBA  :  and  the 
side  EA  (6.  1.)  to  the  side  EB.     In  the  same 
manner,  it  may  be  demonstrated,  that  the  straight 

lines  EC,  ED  are  each  of  them  equal  to  EA,  or  EB  ;  therefore  the  four 
straight  lines  EA,  EB,  EC,  ED,  are  equal  to  one  another;  and  the  circle 
described  from  the  centre  E,  at  the  distance  of  one  of  them,  must  pass 


OF  GEOMETRY.     BOOK  IV.  99 

through  the  extremities  of  the  other  three,  and  be  described  about  the 
square  ABCD. 

PROP.  X.    PROB. 

To  describe  an  isosceles  triangle^  having  each  of  the  angles  at  the  base  double 
of  the  third  angle. 

Take  any  straight  line  AB,  and  divide  (11.  2.)  it  in  the  point  C,  so 
that  the  rectangle  AB.BC  may  be  equal  to  the  square  of  AC  ;  and  from 
the  centre  A,  at  the  distance  AB,  describe  the  circle  BDE,  in  which 
place  (1.  4.)  the  straight  line  BD  equal  to  AC,  which  is  not  greater 
than  the  diameter  of  the  circle  BDE  ;  join  DA,  DC,  and  about  the  tri- 
angle ADC  describe  (5.  4.)  the  circle  ACD  ;  the  triangle  ABD  is  such 
as  is  required,  that  is,  each  of  the  angles  ABD,  ADB  is  double  of  the  an- 
gle BAD. 

Because  the  rectangle  AB.BC  is  equal  to  the  square  of  AC,  and  AC 
equal  to  BD,  the  rectangle  AB.BC  is 
equal  to  the  square  of  BD  ;  and  because 
from  the  point  B  without  the  circle  ACD 
two  straight  lines  BCA,  BD  are  drawn 
to  the  circumference,  one  of  which  cuts, 
and  the  other  meets  the  circle,  and  the 
rectangle  AB.BC  contained  by  the  whole 
of  the  cutting  line,  and  the  part  of  it 
without  the  circle,  is  equal  to  the  square 
of  BD,  which  meets  it ;  the  straight  line 
BD  touches  (37.  3.)  the  circle  ACD. 
And  because  BD  touches  the  circle,  and 
DC  is  drawn  from  the  point  of  contact 
D,  the  angle  BDC  is  equal  (32.  3.)  to  — ^^ 

the  angle  J)h.C  in  the  alternate  segment  •" 

of  the  circle,  to  each  of  these  add  the  angle  CDA ;  therefore  the  whole 
angle  BDA  is  equal  to  the  two  angles  CDA,  DAC  ;  but  the  exterior  angle 
BCD  is  equal  (32.  1.)  to  the  angles  CDA,  DAC  ;  therefore  also  BDA  is 
equal  to  BCD ;  but  BDA  is  equal  (5.  1.)  to  CBD,  because  the  side  AD 
is  equal  to  the  side  AB  ;  therefore  CBD,  or  DBA  is  equal  to  BCD  ;  and 
consequently  the  three  angles  BDA,  DBA,  BCD,  are  equal  to  one  another. 
And  because  the  angle  DBC  is  equal  to  the  angle  BCD,  the  side  BD  is 
equal  (6.  1.)  to  the  side  DC ;  but  BD  was  made  equal  to  CA  ;  therefore 
also  CA  is  equal  to  CD,  and  the  angle  CDA  equal  (5.  1.)  to  the  angle 
DAC ;  therefore  the  angles  CDA,  DAC  together,  are  double  of  the  angle 
DAC  ;  but  BCD  is  equal  to  the  angles  CDA,  DAC  (32.  1.)  ;  therefore 
also  BCD  is  double  of  DAC.  But  BCD  is  equal  to  each  of  the  angles 
BDA,  DBA,  and  therefore  each  of  the  angles  BDA,  DBA,  is  double  of 
the  angle  DAB  ;  wherefore  an  isosceles  triangle  ABD  is  described,  hav- 
ing each  of  the  angles  at  the  base  double  of  the  third  angle. 

"  CoR.  1.  The  angle  BAD  is  the  fifth  part  of  two  right  angles. 
"  For  since  each  of  the  angles  ABD  and  ADB  is  equal  to  twice  the  an- 
"  gle  BAD,  they  are  together  equal  to  four  times  BAD,  and  therefore  all 
"  the  three  angles  ABD,  ADB,  BAD,  taken  together,  are  equal  to  five 


100  ELEMENTS 

"  times  the  angle  BAD.  But  the  three  angles  ABD,  ADB,  BAD  are 
"  equal  to  two  right  angles,  therefore  five  times  the  angle  BAD  is  equal  to 
"  two  right  angles  ;  or  BxiD  is  the  fifth  part  of  two  right  angles." 

"  CoR.  2.  Because  BAD  is  the  fifth  part  of  two,  or  the  tenth  part  of 
"  four  right  angles,  all  the  angles  about  the  centre  A  are  together  equal  to 
"  ten  times  the  angle  BAD,  and  may  therefore  be  divided  into  ten  parts 
"  each  equal  to  BAD.  And  as  these  ten  equal  angles  at  the  centre,  must 
"  stand  on  ten  equal  arcs,  therefore  the  arc  BD  is  one-tenth  of  the  cir- 
"  cumference  ;  and  the  straight  line  BD,  that  is,  AC,  is  therefore  equal  to 
"  the  side  of  an  equilateral  decagon  inscribed  in  the  circle  BDE." 

PROP.  XL     PROB. 

To  inscribe  an  equilateral  and  equiangular  pentagon  in  a  given  circle. 

Let  ABODE  be  the  given  circle,  it  is  required  to  inscribe  an  equilateral 
and  equiangular  pentagon  in  the  circle  ABODE. 

Describe  (10.  4.)  an  isosceles  triangle  FGH,  having  each  of  the  angles 
at  G,  H,  double  of  the  angle  at  F  ;  and  in  the  circle  ABODE  inscribe  (2. 
4.)  the  triangle  AOD  equiangular  to  the  triangle  FGH,  so  that  the  angle 
CAD  be  equal  to  the  angle  at  F,  and  each  of  the  angles  AOD,  ODA  equal 
to  the  angle  at  G  or  H  :  where- 
fore each  of  the  angles  AOD, 
CD  A  is  double  of  the  angle 
CAD.  Bisect  (9.  1.)  the  angles 
AOD,  ODA  by  the  straight  lines 
OE,  DB;  and  join  AB,  BO,  ED, 
EA.  ABODE  is  the  pentagon 
required. 

Because  the  angles  AOD, 
ODA  are  each  of  them  double 
of  CAD,  and  are  bisected  by  the 
straight  lines  CE,  DB,the  five  angles  DAO,  ACE,  EOD,  ODB,  BDA  are 
equal  to  one  another ;  but  equal  angles  stand  upon  equal  arcs  (26.  3.)  ; 
therefore  the  five  arcs  AB,  BO,  CD,  DE,  EA  are  equal  to  one  another  ;  and 
equal  arcs  are  subtended  by  equal  (29.  3.)  straight  lines  ;  therefore  the 
five  straight  lines  AB,  BO,  CD,  DE,  EA  are  equal  to  one  another.  Where- 
fore the  pentagon  ABODE  is  equilateral.  It  is  also  equiangular ;  be- 
cause the  arc  AB  is  equal  to  the  arc  DE  ;  if  to  each  be  added  BOD,  the 
whole  ABOD  is  equal  to  the  whole  EDOB  ;  and  the  angle  AED  stands 
on  the  arc  ABOD,  and  the  angle  BAE  on  the  arc  EDOB  :  therefore  the 
angle  BAE  is  equal  (27.  3.)  to  the  angle  AED  :  for  the  same  reason,  each 
of  the  angles  ABO,  BOD,  ODE  is  equal  to  the  angle  BAE  or  AED  :  there- 
fore the  pentagon  ABODE  is  equiangular;  and  it  has  been  shewn  that  It 
is  equilateral.  Wherefore,  in  the  given  circle,  an  equilateral  and  equian- 
gular pentagon  has  been  inscribed. 

Otherwise. 

"  Divide  the  radius  of  the  given  circle,  so  that  the  rectangle  contained 
"  by  the  whole  and  one  of  the  parts  may  be  equal  to  the  square  of  the  other 


OF  GEOMETRY.    BOOK  IV. 


101 


"(11.  2.).  Apply  in  the  circle,  on  each  side  of  a  given  point,  a  line 
•'  equal  to  the  greater  of  these  parts  ;  then  (2.  Cor.  10.  4.),  each  of  the 
"  arcs  cut  off  will  be  one-tenth  of  the  circumference,  and  therefore  tho 
•'  arc  made  up  of  both  will  be  one-fifth  of  the  circumference  ;  and  if  the 
"  straight  line  subtending  this  arc  be  drawn,  it  will  be  the  side  of  an 
"  equilateral  pentagon  inscribed  in  the  circle." 

PROP.  XII.     PROB. 
To  describe  an  equilateral  and  equiangular  pentagon  about  a  given  circle. 

Let  ABODE  be  the  given  circle,  it  is  required  to  describe  an  equilateral 
and  equiangular  pentagon  about  the  circle  ABODE. 

Let  the  angles  of  a  pentagon,  inscribed  in  the  circle,  by  the  last  pro- 
position, be  in  the  points  A,  B,  0,  D,  E,  so  that  the  arcs  AB,  BC,  CD, 
DE,  EA  are  equal  (11.  4.)  ;  and  through  the  points  A,  B,  C,  D,  E,  draw 
GH,  HK,.KL,  LM,  MG,  touching  (17.  3.)  the  circle  ;  take  the  centre  F, 
and  join  FB,  FK,  FC,  FL,  FD.  And  because  the  straight  line  KL  touch- 
es the  circle  ABODE  in  the  point  0,  to  which  FC  is  drawn  from  the  cen- 
tre F,  FC  is  perpendicular  (18.  3.)  to  KL ;  therefore  each  of  the  angles 
at  0  is  a  right  angle  ;  for  the  same  reason,  the  angles  at  the  points  B,  D  are 
right  angles  ;  and  because  FCKis  aright  angle,  the  square  of  FK  is  equal 
(47.  1.)  to  the  squares  of  FC,  OK.  For  the  same  reason,  the  square  of 
FK  is  equal  to  the  squares  of  FB,  BK  :  therefore  the  squares  of  FC,  CK 
are  equal  to  the  squares  of  FB,  BK,  of  which  the  square  of  FC  is  equal  to 
the  square  of  FB  ;  the  remaining  square  of  CK  is  therefore  equal  to  the 
remaining  square  of  BK,  and  the  straight  line  CK  equal  to  BK  :  and  be- 
cause FB  is  equal  to  FC,  and  FK  common  to  the  triangles  BFK,  CFK, 
the  two  BF,  FK  are  equal  to  the  two  OF,  FK  ;  and  the  base  BK  is  equal 
to  the  base  KC  ;  therefore  the  angle  BFK  is  equal  (8.  1.)  to  the  angle 
KFC,  and  the  angle  BKF  to  FKC  ;  wherefore  the  angle  BFC  is  double 
of  the  angle  KFC,  and  BKC  double  of  FKC  :  for  the  same  reason,  the  an- 
gle CFD  is  double  of  the  angle  CFL,  and  OLD  double  of  OLF  :  and  be- 
cause the  arc  BC  is  equal  to  the  arc  CD,  the  angle  BFC  is  equal  (27.  3.) 
to  the  angle  CFD  :  and  BFC  is  double  of  the  angle  KFC,  and  CFD 
double  of  CFL  ;  therefore  the  angle 
KFC  is  equal  to  the  angle  CFL  : 
now  the  right  angle  FCK  is  equal  to 
the  right  angle  FOL  ;  and  therefore, 
in  the  two  triangles  FKb,  FLO,  there 
are  two  angles  of  one  equal  to  two  an- 
gles of  the  other,  each  to  each,  and  the 
side  FC,  which  is  adjacent  to  the 
equal  angles  in  each,  is  common  to 
both  ;  therefore  the  other  sides  are 
equal  (26. 1 .)  to  the  other  sides,and  the 
third  angle  to  the  third  angle  ;  there- 
fore the  straight  line  KC  is  equal  to 
CL,  and  the  angle  FKC  to  the  angle 
FLO  :  and  because  KC  is  equal  to  CL,  KL  is  double  of  KC  ;  in  the  same 
manner,  it  may  be  shewn  that  HK  is  double  of  BK ;  and  because  BK  is 


102  '  ELEMENTS 

equal  to  KC,  as  was  demonstrated,  and  KL  is  double  of  KC,  and  HK  double 
of  BK,  HK  is  equal  to-KL  ;  in  like  manner,  it  may  be  shewn  that  GH,  GM, 
ML  are  each  of  them  equal  to  HK  or  KL :  therefore  the  pentagon  GHKLM 
is  equilateral.  It  is  also  equiangular  ;  for,  since  the  angle  FKC  is  equal  to 
the  angle  FLC,  and  the  angle  HKL  double  of  the  angle  FKC,  and  KLM 
double  of  FLC,  as  was  before  demonstrated,  the  angle  HKL  is  equal  to 
KLM  ;  and  in  like  manner  it  may  be  shewn,  that  each  of  the  angles  KHG, 
HGM,  GML  is  equal  to  the  angle  HKL  or  KLM  ;  therefore  the  five  an- 
gles GHK,  HKL,  KLM,  LMG,  MGH  being  equal  to  one  another,  the  pen- 
tagon GHKLM  is  equiangular  ;  and  it  is  equilateral  as  was  demonstra- 
ted :  and  it  is  described  about  the  circle  ABCDE. 

PROP.  XHL     PROB. 

To  inscribe  a  circle  in  a  given  equilateral  and  equiangular  pentagon. 

Let  ABCDE  be  the  given  equilateral  and  equiangular  pentagon  ;  it  is 
required  to  inscribe  a  circle  in  the  pentagon  ABCDE. 

Bisect  (9. 1.)  the  angles  BCD,  CDE  by  the  straight  lines  CF,  DF,  and 
from  the  point  F,  in  which  they  meet,  draw  the  straight  lines  FB,  FA, 
FE  ;  therefore,  since  BC  is  equal  to  CD,  and  CF  common  to  the  trian- 
gles BCF,  DCF,  the  two  sides  BC,  CF  are  equal  to  the  two  DC,  CF ; 
and  the  angle  BCF  is  equal  to  the  angle  DCF :  therefore  the  base  BF  is 
equal  (4.  1.)  to  the  base  FD,  and  the  other  angles  to  the  other  angles,  to 
which  the  equal  sides  are  opposite  ;  therefore  the  angle  CBF  is  equal,  to 
the  angle  CDF  :  and  because  the  angle  CDE  is  double  of  CDF,  and  CDE 
equal  to  CBA,  and  CDF  to  CBF  ;  CBA  is  also  double  of  the  angle  CBF  ; 
therefore  the  angle  ABF  is  equal  to  the 
angle  CBF  ;  wherefore  the  angle  ABC 
is  bisected  by  the  straight  line  BF  :  in 
the  same  manner,  it  may  be  demonstra- 
ted that  the  angles  BAE,  AED,  are  bi- 
sected by  the  straight  lines  AF,  EF  : 
from  the  point  F  draw  (12.  1.)  FG, 
FH,  FK,  FL,  FM  perpendiculars  to 
the  straight  lines  AB,  BC,  CD,  DE, 
EA ;  and  because  the  angle  HCF  is 
equal  to  KCF,  and  the  right  angle 
FHC  equal  to  the  right  angle  FKC  ;  in 
the  triangles  FHC,  FKC  there  are  two 
angles  of  one  equal  to  two  angles  of  the  other,  and  the  side  FC,  which  is 
opposite  to  one  of  the  equal  angles  in  each,  is  common  to  both  ;  therefore, 
the  other  sides  shall  be  equal  (26.  1.),  each  to  each ;  wherefore  the  per- 
pendicular FH  is  equal  to  the  perpendicular  FK  :  in  the  same  manner  it 
may  be  demonstrated,  that  FL,  FM,  FG  are  each  of  them  equal  to  FH,  or 
FK;  therefore  the  five  straight  lines  FG,  FH,  FK,  FL,  FM  are  equal  to 
one  another  ;  wherefore  the  circle  described  from  the  centre  F,  at  the  dis- 
tance of  one  of  these  five,  will  pass  through  the  extremities  of  the  other 
four,  and  touch  the  straight  lines  AB,  BC,  CD,  DE,  E A,  because  that  the 
angles  at  the  points  G,  H,  K,  L,  M  are  right  angles,  and  that  a  straight  line 
drawn  from  the  extremity  of  the  diameter  of  a  circle  at  right  angles  to  it, 


OF  GEOMETRY.     BOOK  IV.  103 

touches  (1.  Cor.  16.  3.)  the  circle  ;  therefore  each  of  the  straight  lines  AB, 
BC,  CD,  DE,  EA  touches  the  circle  ;  wherefore  the  circle  is  inscribed  in 
the  pentagon  ABODE. 

PROP.  XIV.     PROB. 
To  describe  a  circle  about  a  given  equilateral  and  equiangular  pentagon. 

Let  ABODE  be  the  given  equilateral  and  equiangular  pentagon  ;  it  is 
required  to  describe  a  circle  about  it. 

Bisect  (9.  1.)  the  angles  BCD,  ODE  by  the  straight  lines  OF,  FD,  and 
from  the  point  F,  in  which  they  meet,  draw 
the  straight  lines  FB,  FA,  FE  to  the  points 
B,  A,  E.  It  may  be  demonstrated,  in  the 
same  manner  as  in  the  preceding  proposition, 
that  the  angles  CBA,  BAE,  AED  are  bisect- 
ed by  the  straight  lines  FB,  FA,  FE  :  and 
because  that  the  angle  BCD  is  equal  to  the 
angle  ODE,  and  that  FCD  is  the  half  of  the 
angle  BCD,  and  CDF  the  half  of  ODE  ;  the 
angle  FCD  is  equal  to  FDC  ;  wherefore  the 
side  OF  is  equal  (6.  1 .)  to  the  side  FD  :  in 
like  manner  it  may  be  demonstrated,  that  FB, 
FA,  FE  are  each  of  them  equal  to  FC,  or  FD  :  therefore  the  five  straight 
lines  FA,  FB,  FC,  FD,  FE  are  equal  to  one  another ;  and  the  circle  de- 
scribed from  the  centre  F,  at  the  distance  of  one  of  them,  will  pass  through 
the  extremities  of  the  other  four,  and  be  described  about  the  equilateral 
and  equiangular  pentagon  ABODE. 

PROP.  XV.     PROB. 

To  inscribe  an  equilateral  and  equiangular  hexagon  in  a  given  circle. 

Let  ABCDEF  be  the  given  circle ;  it  is  required  to  inscribe  an  equi- 
lateral and  equiangular  hexagon  in  it. 

Find  the  centre  G  of  the  circle  ABCDEF,  and  draw  the  diameter  AGD  : 
and  from  D,  as  a  centre,  at  the  distance  DG,  describe  the  circle  EGCH, 
join  EG,  CG,  and  produce  them  to  the  points  B,  F  ;  and  join  x\B,  BC, 
CD,  DE,  EF,  FA  :  the  hexagon  ABCDEF  is  equilateral  and  equiangular. 

Because  G  is  the  centre  of  the  circle  ABCDEF,  GE  is  equal  to  GD : 
and  because  D  is  the  centre  of  the  circle  EGCH,  DE  is  equal  to  DG ; 
wherefore  GE  is  equal  to  ED,  and  the  triangle  EGD  is  equilateral;  and 
therefore  its  three  angles  EGD,  GDE,  DEG  are  equal  to  one  another 
(Cor.  5.  1.) ;  and  the  three  angles  of  a  triangle  are  equal  (32.  1.)  to  two 
right  angles ;  therefore  the  angle  EGD  is  the  third  part  of  two  right  an- 
gles :  in  the  same  manner  it  may  be  demonstrated  that  the  angle  DGC  is 
also  the  third  part  of  two  right  angles  ;  and  because  the  straight  line  GO 
makes  with  EB  the  adjacent  angles  EGO,  CGB  equal  (13.  1.)  to  two 
right  angles  ;  the  remaining  angle  CGB  is  the  third  part  of  two  right 
angles  ;  therefore  the  angles  EGD,  DGC,  CGB,  are  equal  to  one  an- 
other; and  also  the  angles  vertical  to  them,  BGA,  AGF,  FGE  (15. 


104 


ELEMENTS 


1.);  therefore  the  six  angles  EGD,  DGC, 
CGB,  BGA,  AGF,  FGE  are  equal  to  one  an- 
other. But  equal  angles  at  the  centre  stand 
upon  equal  arcs  (26.  3.) :  therefore  the  six 
arcs  AB,  BC,  CD,  DE,  EF,  FA  are  equal 
to  one  another  :  and  equal  arcs  are  subtend- 
ed by  equal  (29.  3.)  straight  lines  ;  there- 
fore the  six  straight  lines  are  equal  to  one 
another,  and  the  hexagon  ABCDEF  is 
equilateral.  It  is  also  equiangular;  for, 
since  the  arc  AF  is  equal  to  ED,  to  each  of 
these  add  the  arc  ABCD  ;  therefore  the 
whole  arc  FABCD  shall  be  equal  to  the 
whole  EDCBA:  and  the  angle  FED  stands 
upon  the  arc  FABCD,  and  the  angle  AFE 
upon  EDCBA;  therefore  the  angle  AFE 
is  equal  to  FED  :  in  the  same  manner  it  may  be  demonstrated,  that  the 
other  angles  of  the  hexagon  ABCDEF  are  each  of  them  equal  to  the 
angle  AFE  or  FED  ;  therefore  the  hexagon  is  equiangular ;  it  is  also 
equilateral,  as  was  shown  ;  and  it  is  inscribed  in  the  given  circle  ABCDEF. 

Cor.  From  this  it  is  manifest,  that  the  side  of  the  hexagon  is  equal  to 
the  straight  line  from  the  centre,  that  is,  to  the  radius  of  the  circle. 

And  if  through  the  points  A,  B,  C,  D,  E,  F,  there  be  drawn  straight 
lines  touching  the  circl^,  an  equilateral  and  equiangular  hexagon  shall  be 
described  about  it,  which  may  be  demonstrated  from  what  has  been  said 
of  the  pentagon  ;  and  likewise  a  circle  may  be  inscribed  in  a  given  equi- 
lateral and  equiangular  hexagon,  and  circumscribed  about  it,  by  a  method 
like  to  that  used  for  the  pentagon. 


PROP.  XVI.     PROB. 


To  inscribe  an  equilateral  and  equiangular  qumdecagon  in  a  given 

circle. 

Let  ABCD  be  the  given  circle ;  it  is  required  to  inscribe  an  equilateral 
and  equiangular  quindecagon  in  the  circle  ABCD. 

Let  AC  be  the  side  of  an  equilateral  triangle  inscribed  (2.  4.)  in  the 
circle,  and  AB  the  side  of  an  equilateral 
and  equiangular  pentagon  inscribed  (11.  4.) 
in  the  same  ;  therefore,  of  such  equal  parts 
as  the  whole  circumference  ABCDF  con- 
tains fifteen,  the  arc  ABC,  being  the  third 
part  of  the  whole,  contains  five ;  and  the 
arc  AB,  which  is  the  fifth  part  of  the  whole, 
contains  three ;  therefore  BC  their  differ- 
ence contains  two  of  the  same  parts :  bi- 
sect (30.  3.)  BC  in  E  ;  therefore  BE,  EC 
are,  each  of  them,  the  fifteenth  part  of  the 
whole  circumference  ABCD  :  therefore,  if 
the  straight  lines  BE,  EC  be  drawn,  and 


OF  GEOMETRY.     BOOK  IV.  105 

straight  lines  equal  to  them  be  placed  (1.  4.)  around  in  the  whole  circle, 
an  equilateral  and  equiangular  quindecagon  will  be  inscribed  in  it. 

And  in  the  same  manner  as  was  done  in  the  pentagon,  if  through  the 
points  of  division  made  by  inscribing  the  quindecagon,  straight  lines  be 
drawn  touching  the  circle,  an  equilateral  and  equiangular  quindecagon  may- 
be described,  about  it :  and  likewise,  as  in  the  pentagon,  a  circle  may  be 
inscribed  in  a  given  equilateral  and  equiangular  quindecagon,  and  cir* 
cumscribed  about  it. 

SCHOLIUM. 

Any  regular  polygon  being  inscribed,  if  the  arcs  subtended  by  its  sides 
be  severally  bisected,  the  chords  of  those  semi-arcs  will  form  a  new  regu- 
lar polygon  of  double  the  number  of  sides  :  thus,  from  having  an  inscribed 
square,  we  may  inscribe  in  succession  polygons  of  8,  16,  32,  64,  &c.  sides  ; 
from  the  hexagon  may  be  formed  polygons  of  12,  24,  48,  96,  &c.  sides ; 
from  the  decagon  polygons  of  20,  40,  80,  &c.  sides  ;  and  from  the  pente- 
decagon  we  may  inscribe  polygons  of  30,  60,  &c.  sides ;  and  it  is  plain 
that  each  polygon  will  exceed  the  preceding  in  surface  or  area. 

It  is  obvious  that  any  regular  polygon  whatever  might  be  inscribed  in  a 
circle,  provided  that  its  circumference  could  be  divided  into  any  proposed 
number  of  equal  parts  ;  but  such  division  of  the  circumference  like  the  tri- 
section  of  an  angle,  which  indeed  depends  on  it,  is  a  problem  which  has 
not  yet  been  effected.  There  are  no  means  of  inscribing  in  a  circle  a  regu- 
lar heptagon,  or  which  is  the  same  thing,  the  circumference  of  a  circle  can- 
not be  divided  into  seven  equal  parts,  by  any  method  hitherto  discovered. 

It  was  long  supposed,  that  besides  the  polygons  above  mentioned,  no 
other  could  be  inscribed  by  the  operations  of  elementary  Geometry,  or, 
what  amounts  to  the  same  thing,  by  the  resolution  of  equations  of  the  first 
and  second  degree.  But  M.  Gauss,  of  Gottingen,  at  length  proved,  in  a 
work  entitled  Disquisitiones  Arithmeticcs,  Lipsie,  1801,  that  the  circumfer- 
ence of  a  circle  could  be  divided  into  any  number  of  equal  parts,  capable 
of  being  expressed  by  the  formula  2"4-l,  provided  it  be  a  prime  number, 
that  is,  a  number  that  cannot  be  resolved  into  factors. 

The  number  3  is  the  simplest  of  this  kind,  it  being  the  value  of  the 
above  formula  when  n=:l  ;  the  next  prime  number  is  5,  and  this  is  also 
contained  in  the  formula  ;  that  is,  when  nz=2.  But  polygons  of  3  and  5 
sides  have  already  been  inscribed.  The  next  prime  number  expressed  by 
the  formula  is  17  ;  so  that  it  is  possible  to  inscribe  a  regular  polygon  of 
17  sides  in  a  circle. 

For  the  investigation  of  Gauss's  theorem,  which  depends  upon  the  the- 
ory of  algebraical  equations,  the  student  may  consult  Barlow's  Theory  of 
Numbers. 

14 


ELEMENTS 


GEOMETRY 


BOOK  V. 


In  the  demonstrations  of  this  book  there  are  certain  "  signs  or  characters" 
which  it  has  been  found  convenient  to  employ. 

*  1.  The  letters  A,  B,  C,  &;c.  are  used  to  denote  magnitudes  of  any  kind. 

"The  letters  m,  n,  p,  q,  are  used  to  denote  numbers  only. 

It  is  to  be  observed,  that  in  speaking  of  the  magnitudes  A,  B,  C,  &c., 
we  mean,  in  reality,  those  which  these  letters  are  employed  to  repre- 
sent ;  they  may  be  either  lines,  surfaces,  or  solids. 

"  2.  When  a  mimber,  or  a  letter  denoting  a  number,  is  written  close  to 
"  another  letter  denoting  a  magnitude  of  any  kind,  it  signifies  that  the 
"  magnitude  is  multiplied  by  the  number.  Thus,  3A  signifies  three 
"  times  A;  mB,  m  times  B,  or  a  multiple  of  B  bym.  When  the  num- 
"  her  is  intended  to  multiply  two  or  more  magnitudes  that  follow,  it  is 
"  written  thus,  wi(A+B),  which  signifies  the  sum  of  A  and  B  taken  m 
"times  ;  »?(A— B)  is  m  times  the  excess  of  A  above  B. 

"  Also,  when  two  letters  that  denote  numbers  are  written  close  to  one  an- 
"  other,  they  denote  the  product  of  those  numbers,  when  multiplied  into 
"  one  another.  Thus,  mn  is  the  product  of  m  into  n  ;  and  mnA  is  A  mul- 
"  tiplied  by  the  product  of  m  into  n. 

DEFINITIONS. 

1.  A  less  magnitude  is  said  to  be  a  part  of  a  greater  magnitude,  when  the 
less  measures  the  greater,  that  is,  when  the  less  is  contained  a  certain 
number  of  times,  exactly,  in  the  greater. 

2.  A  greater  magnitude  is  said  to  be  a  multiple  of  a  less,  when  the  greater 
is  measured  by  the  less,  that  is,  when  the  greater  contains  the  less  a  cer- 
tain number  of  times  exactly. 

3.  Ratio  is  a  mutual  relation  of  two  magnitudes,  of  the  same  kind,  to  one 
another,  in  respect  of  quantity. 


OF  GEOMETRY.    BOOK  V.  107 

4.  Magnitudes  are  said  to  be  of  tlie  same  kind,  when  the  less  can  be  mul- 
tiplied so  as  to  exceed  the  greater  ;  and  it  is  only  such  magnitudes  that 
are  said  to  have  a  ratio  to  one  another. 

5.  If  there  be  four  magnitudes,  and  if  any  equimultiples  whatsoever  be 
taken  of  the  first  and  third,  and  any  equimultiples  whatsoever  of  the  se- 
cond and  fourth,  and  if,  according  as  the  multiple  of  the  first  is  greater 
than  the  multiple  of  the  second,  equal  to  it,  or  less,  the  multiple  of  the 
third  is  also  greater  than  the  multiple  of  the  fourth,  equal  to  it,  or  less  ; 
then  the  first  of  the  magnitudes  is  said  to  have  to  the  second  the  same 
ratio  that  the  third  has  to  the  fourth. 

6.  Magnitudes  are  said  to  be  proportionals,  when  the  first  has  the  same 
ratio  to  the  second  that  the  third  has  to  the  fourth  ;  and  the  third  to  the 
fourth  the  same  ratio  which  the  fifth  has  to  the  sixth,  and  so  on  whatever 
be  their  number. 

"  When  four  magnitudes.  A,  B,  C,  D  are  proportionals,  it  is  usual  to  say 
"  that  A  is  to  B  as  C  to  D,  and  to  write  them  thus,  A  :  B ;:  C  :  D,  or 
"thus,  A  :  B=C  :  D." 

7.  When  of  the  equimultiples  of  four  magnitudes,  taken  as  in  the  fifth 
definition,  the  multiple  of  the  first  is  greater  than  that  of  the  second, 
but  the  multiple  of  the  third  is  not  greater  than  the  multiple  of  the  fourth  : 
then  the  first  is  said  to  have  to  the  second  a  greater  ratio  than  the  third 
magnitude  has  to  the  fourth  :  and,  on  the  contrary,  the  third  is  said  to 
have  to  the  fourth  a  less  ratio  than  the  first  has  to  the  second. 

8.  When  there  is  any  number  of  magnitudes  greater  than  two,  of  which 
the  first  has  to  the  second  the  same  ratio  that  the  second  has  to  the 
third,  and  the  second  to  the  third  the  same  ratio  which  the  third  has  to 
the  fourth,  and  so  on,  the  magnitudes  are  said  to  be  continual  propor- 
tionals. 

9.  When  three  magnitudes  are  continual  proportionals,  the  second  is  said' 
to  be  a  mean  proportional  between  the  other  two. 

10.  When  there  is  any  number  of  magnitudes  of  the  same  kind,  the  first 
is  said  to  have  to  the  last  the  ratio  compounded  of  the  ratio  which  the 
first  has  to  the  second,  and  of  the  ratio  which  the  second  has  to  the 
third,  and  of  the  ratio  which  the  third  has  to  the  fourth,  and  so  on  unto 
the  last  magnitude. 

For  example,  if  A,  B,  C,  D,  be  four  magnitudes  of  the  same  kind,  the 
first  A  is  said  to  have  to  the  last  D,  the  ratio  compounded  of  the  ratio 
of  A  to  B,  and  of  the  ratio  of  B  to  C,  and  of  the  ratio  of  C  to  D ;  or, 
the  ratio  of  A  to  D  is  said  to  be  compounded  of  the  ratios  of  A  to  B, 
B  to  C,  and  C  to  D. 

And  if  A  :  B ::  E  :  F  ;  and  B  :  C ::  G  :  H,  and  C  :  D ::  K  :  L,  then,  since 
by  this  definition  A  has  to  D  the  ratio  compounded  of  the  ratios  of  A  to 
B,  B  to  C,  C  to  D ;  A  may  also  be  said  to  have  to  D  the  ratio  compounded 
of  the  ratios  which  are  the  same  with  the  ratios  of  E  to  F,  G  to  H, 
and  K  to  L. 


108  ELEMENTS 

In  like  manner,  the  same  things  being  supposed,  if  M  has  to  N  the  same 
ratio  which  A  has  to  D,  then,  for  shortness'  sake,  M  is  said  to  have  to 
N  a  ratio  compounded  of  the  same  ratios  which  compound  the  ratio  of 
i^  to  D  ;  that  is,  a  ratio  compounded  of  the  ratios  of  E  to  F,  G  to  H, 
and  K  to  L. 

11.  If  three  magnitudes  are  continual  proportionals,  the  ratio  of  the  first 
to  the  third  is  said  to  be  duplicate  of  the  ratio  of  the  first  to  the  second 

"  Thus,  if  A  be  to  B  as  B  to  C,  the  ratio  of  A  to  C  is  said  to  be  duplicate 
"  of  the  ratio  of  A  to  B.  Hence,  since  by  the  last  definition,  the  ratio 
"  of  A  to  C  is  compounded  of  the  ratios  of  A  to  B,  and  B  to  C,  a  ratio, 
"  which  is  compounded  of  two  equal  ratios,  is  duplicate  of  either  of 
"  these  ratios." 

12.  If  four  magnitudes  are  continual  proportionals,  the  ratio  of  the  first 
to  the  fourth  is  said  to  be  triplicate  of  the  ratio  of  the  first  to  the  second, 
or  of  the  ratio  of  the  second  to  the  third,  &c. 

"  So  also,  if  there  are  five  continual  proportionals  ;  the  ratio  of  the  first 
"  to  the  fifth  is  called  quadruplicate  of  the  ratio  of  the  first  to  the  se- 
"cond  ;  and  so  on,  according  to  the  number  of  ratios.  Hence,  a  ratio 
"  compounded  of  three  equal  ratios,  is  triplicate  of  any  one  of  those  ra- 
"  tios  ;  a  ratio  compounded  of  four  equal  ratios  quadruplicate,"  &c. 

13.  In  proportionals,  the  antecedent  terms  are  called  homologous  to  one 
another,  as  also  the  consequents  to  one  another. 

Geometers  make  use  of  the  following  technical  words  to  signify  certain 
ways  of  changing  either  the  order  or  magnitude  of  proportionals,  so  as 
that  they  contmue  still  to  be  proportionals. 

14.  Permutando,  or  alternando,  by  permutation,  or  alternately  ;  this  word 
is  used  when  there  are  four  proportionals,  and  it  is  inferred,  that  the  first 
has  the  same  ratio  to  the  third  which  the  second  has  to  the  fourth ;  or 
that  the  first  is  to  the  third  as  the  second  to  the  fourth  :  See  Prop.  16. 
of  this  Book. 

15.  Invertendo,  by  inversion  :  When  there  are  four  proportionals,  and  it  is 
inferred,  that  the  second  is  to  the  first,  as  the  fourth  to  the  third.  Prop 
A.  Book  5. 

16.  Componendo,  by  composition :  When  there  are  four  proportionals,  and 
it  is  inferred,  that  the  first,  together  with  the  second,  is  to  the  second  as 
the  third,  together  with  the  fourth,  is  to  the  fourth.     18th  Prop.  Book  5. 

17.  Dividendo,  by  division ;  when  there  are  four  proportionals,  and  it  is 
inferred  that -the  excess  of  the  first  above  the  second,  is  to  the  second, 
as  the  excess  of  the  third  above  the  fourth,  is  to  the  fourth.  1 7th  Prop. 
Book  5. 

18.  Convertendo,  by  conversion  ;  when  there  are  four  proportionals,  and 
it  is  inferred,  that  the  first  is  to  its  excess  above  the  second,  as  the  third 
to  its  excess  above  the  fourth.     Prop.  D.  Book  5. 


OF  GEOMETRY.    BOOK  V.  109 

19.  Ex  sequali  (sc.  distantia),  or  ex  asquo,  from  equality  of  distance  ; 
when  there  is  any  number  of  magnitudes  more  than  two,  and  as  many 
others,  so  that  they  are  proportionals  when  taken  two  and  two  of  each 
rank,  and  it  is  inferred,  that  the  first  is  to  the  last  of  the  first  rank  of 
magnitudes,  as  the  first  is  to  the  last  of  the  others  ;  Of  this  there  are  the 
two  following  kinds,  which  arise  from  the  difierent  order  in  which  the 
magnitudes  are  taken  two  and  two. 

20.  Ex  sequali,  from  equality ;  this  term  is  used  simply  by  itself,  when 
the  first  magnitude  is  to  the  second  of  the  first  rank,  as  the  first  to  the 
second  of  the  other  rank ;  and  as  the  second  is  to  the  third  of  the  first 
rank,  so  is  the  second  to  the  third  of  the  other  ;  and  so  on  in  order,  and . 
the  inference  is  as  mentioned  in  the  preceding  definition  ;  whence  this 
is  called  ordinate  proportion. 

It  is  demonstrated  in  the  22d  Prop.  Book  5. 

21.  Ex  aequali,  in  proportione  perturbata,  sen  inordinata  :  from  equality,  in 
perturbate,  or  disorderly  proportion ;  this  term  is  used  when  the  first 
magnitude  is  to  the  second  of  the  first  rank,  as  the  last  but  one  is  to  the 
last  of  the  second  rank  ;  and  as  the  second  is  to  the  third  of  the  first 
rank,  so  is  the  last  but  two  to  the  last  but  one  of  the  second  rank ;  and 
as  the  third  is  to  the  fourth  of  the  first  rank,  so  is  the  third  from  the  last, 
to  the  last  but  two,  of  the  second  rank ;  and  so  on  in  a  cross,  or  inverse^ 
order ;  and  the  inference  is  as  in  the  19th  definition.  It  is  demonstrated 
in  the  23d  Prop,  of  Book  5. 

AXIOMS. 

1.  Equimultiples  of  the  same,  or  of  equal  magnitudes,  are  equal  to  one 
another. 

2.  Those  magnitudes  of  which  the  same,  or  equal  magnitudes,  are  equi- 
multiples, are  equal  to  one  another. 

3.  A  multiple  of  a  greater  magnitude  is  greater  than  the  same  multiple  of 
a  less. 

4.  That  magnitude  of  which  a  multiple  is  greater  than  the  same  multi- 
ple of  another,  is  greater  than  that  other  magnitude. 

PROP.  I.     THEOR. 

If  any  number  of  magnitudes  he  equimultiples  of  as  many  others,  each  of 
each,  what  multiple  soever  any  one  of  the  first  is  of  its  part,  the  same  muU 
tiple  is  the  sum  of  all  the  first  of  the  sum  of  all  the  rest. 

Let  any  number  of  magnitudes  A,  B,  and  C  be  equimultiples  of  as  many 
others,  D,  E,  and  F,  each  to  each,  A-f  B  +  C  is  the  same  multiple  of  0+ 
E  +  F,  that  A  is  of  D. 

Let  A  contain  D,  B  contain  E,  and  C  contain  F,  each  the  same  number 
of  times,  as,  for  instance,  three  times 


no  ELEMENTS 

Then,  because  A  contains  D  three  times,  -  A=D+D4-D. 

For  the  same  reason,  %  B=E4-E-i-E  ; 

And  also,  C=F4-F+F. 

Therefore,  adding  equals  to  equals  (Ax.  2.  1.),  A4-B-|-C  is  equal  to 
D-fE-f-F,  taken  three  times.  In  the  same  manner,  if  A,  B,  and  C  were 
each  any  other  equimultiple  of  D,  E,  and  F,  it  would  be  shown  that  A+ 
B  +  C  was  the  same  multiple  of  D  +  E  +  F. 

CoR.  Hence,  if  m  be  any  number,  mD4-'wE-4-^F=m(D-4-E+F). 
For  mD,  mE,  and  mF  are  multiples  of  D,  E,  and  F  by  m,  therefore  their 
sum  is  also  a  multiple  of  D+E-f-F  by  m. 

PROP.  II.     THEOR. 

If  to  a  multiple  of  a  magnitude  hy  any  number,  a  multiple  of  the  same  mag' 
nitude  hy  any  numher  he  added,  the  sum  will  be  the  same  multiple  of  that 
magnitude  that  the  sum  of  the  two  numbers  is  of  unity. 

Let  A=mC,  and  B=nC  ;  A+B=(m4-n)C. 

For,  since  A=mC,  A=C4-C4-C+&:c.  C  being  repeated m times.  For 
the  same  reason,  B=C+C+<^c.  C  being  repeated  n  times.  Therefore, 
adding  equals  to  equals,  A-{-B  is  equal  to  C  taken  m-{-n  times ;  that  is, 
A^-B=(m+n^C.  Therefore  A+B  contains  C  as  oft  as  there  are  units 
inm+n. 

Cor.  1.  In  the  same  way,  if  there  be  any  number  of  multiples  what- 
soever, as  A=»iE,  B=nE,  C=j9E,  it  is  shown,  that  A+B4-C=(7»-f  h 
+;,)E. 

Cor.  2.  Hence  also,  since  A4-B  +  C=(m-|-n+/))E,andsinceA=mE, 
B=7iE,  and  C=pE,  ?nE+nE4TpE=(w+n+i>)E. 

PROP.  III.     THEOR. 

If  the  first' of  three  magnitudes  contain  the  second  as  often  as  there  are  units 
in  a  certain  numher,  and  if  the  second  contain  the  third  also,  as  often  as 
there  are  units  in  a  certain  numher,  the  first  will  contain  the  third  as  often 
as  there  are  units  in  the  product  of  these  two  numbers. 

Let  A=mB,  and  B=nC  ;  then  A=mnC. 

Since  B=nC,  mB=nC-{-nC-{-iLc.  repeated  m  times.  But  nC-\-nC, 
Sic.  repeated  m  times  is  equal  to  C  (2.  Cor.  2.  5.).  multiplied  by  n+w-f&c. 
n  being  added  to  itself  m  times  ;  but  n  added  to  itself  m  times,  is  n  multi- 
plied by  m,  or  mn.  Therefore  nC-^-nC-^&c.  repeated  m  times=m«C ; 
whence  also  mB=mnC,  and  by  hypothesis  A=mB,  therefore  A=mnC 


OF  GEOMETRY.    BOOK  V.  Ill 


PROP.  IV.    THEOR. 

If  the  first  of  four  magnitudes  has  the  same  ratio  to  the  second  which  the  third 
has  to  the  fourth  J  and  if  any  equimultiples  whatever  he  taken  of  the  first  and 
third,  and  any  whatever  of  the  second  and  fourth ;  the  multiple  of  the  first 
shall  have  the  same  ratio  to  the  multiple  of  the  second,  that  the  multiple  of 
the  third  has  to  the  multiple  of  the  fourth. 

Let  A  :  B  :  :  C  :  D,  and  let  m  and  n  be  any  two  numbers ;  mA  :  nB  :  : 
mC  :  nD. 

Take  of  mA  and  mC  equimultiples  by  any  number  j9,  and  of  nB  and  nD 
equimultiples  by  any  number  q.  Then  the  equimultiples  of  mk,  and  mC 
by  p,  are  equimultiples  also  of  A  and  0,  for  they  contain  A  and  C  as  oft  as 
there  are  units  inpm  (3.  5.),  and  are  equal  to  pmk  and  pmC  For  the  same 
reason  the  multiples  of  wB  and  nD  by  q,  are  qnE,  qnD.  Since,  therefore, 
A  :  B  : :  0  :  D,and  of  A  and  0  there  are  taken  any  equimultiples,  y'lz.pmA. 
andpmC,  and  of  B  and  D,  any  equimultiples  ^nB,  qnT),  if  pmA  be  greater 
than  qnB,pmC  must  be  greater  than  qnD  (def.  5.  5.) ;  if  equal,  equal ;  and 
if  less,  less.  But  pmA,  pmC  are  also  equimultiples  of  mA.  and  mC,  and 
qnBj  qnJ)  are  equimultiples  of  nB  and  nD,  therefore  (def.  5.  5.),  mA  :  nB 
: :  mC  :  nD. 

Cor.  In  the  same  manner  it  may  be  demonstrated,  that  if  A  :  B  : :  C  : 
D,  and  of  A  and  C  equimultiples  be  taken  by  any  number  «i,  viz.  mA  and 
mC,  mA  :  B  : :  mC  :  D.  This  may  also  be  considered  as  included  in  the 
proposition,  and  as  being  the  case  when  n=:l. 

PROP.  V.     THEOR. 

If  one  magnitude  he  the  same  multiple  of  another,  which  a  magnitude  taken 
from  the  first  is  of  a  magnitude  taken  from  the  other  ;  the  remainder  is  the 
same  multiple  of  the  remainder,  that  the  whole  is  of  the  whole. 

Let  mA  and  mB  be  any  equimultiples  of  the  two  magnitudes  A  and  B, 
of  which  A  is  greater  than  B  ;  mA — mB  is  the  same  multiple  of  A — B 
that  mA  is  of  A,  that  is,  mA~mB=m(A— B). 

Let  D  be  the  excess  of  A  above  B,  then  A— B=D,  and  adding  B  to 
both,  A=D+B.  Therefore  (1.  5.)  mA=mD4-'"B  ;  take  mB  from  both, 
and  mA-— mB=mD  ;  but  D=A--B,  therefore  mA--mB=m(A— B). 

PROP.  VI.     THEOR. 

If  from  a  multiple  of  a  magnitude  hy  any  number  a  multiple  of  the  same  mag- 
nitude by  a  less  number  be  taken  away,  the  remainder  will  be  the  same  mul- 
tiple of  that  magnitude  that  the  difference  of  the  numbers  is  of  unity. 

Let  mA  and  nA  be  multiples  of  the  magnitude  A,  by  the  numbers  m  and 
n,  and  let  m  be  greater  than  n  ;  mA — nA  contains  A  as  oft  as  m — n  con- 
tains unity,  or  mA— nA=:(m — n)A. 


112  ELEMENTS 

Let  m — n-=q',  then  m^m-^-q.  Therefore  (2.  5.)  mA.=nk-\-qk  ;  take 
nA  from  both,  and  wiA — nA=qA.  Therefore  mA—nA  contains  Aas  oft 
as  there  are  units  in  q,  that  is,  in  m—n,  or  ttiA— nA  =  (m— 7z)A. 

CoR.  When  the  difference  of  the  two  numbers  is  equal  to  unity  or  m— > 
n=l,  then  mA— nA=A. 

PROP.  A.    THEOR. 

If  four  magnitudes  be  proportionals,  they  are  proportionals  also  when  taken 

inversely. 

If  A  :  B  : :  C  :  D,  then  also  B  :  A  : :  D  :  C. 

Let  mA  and  mO  be  any  equimultiples  of  A  and  C  ;  nB  and  nD  any  equi- 
multiples of  B  and  D.  Then,  because  A  :  B  : :  C  :  D,  if  mA  be  less  than 
nB,  mC  will  be  less  than  nD  (def.  5.  5.),  that  is,  if  wB  be  greater  than  mA, 
nD  will  be  greater  than  mC.  For  the  same  reason,  if  nB=mA,  nD=mC, 
and  if  wB /mA,  nD/ mC.  But  nB,  nD  are  any  equimultiples  of  B  and  D, 
and  mA,  mC  any  equimultiples  of  A  and  C,  therefore  (def.  5.  5.),  B  :  A  • . 
D:C. 

PROP.  B.    THEOR. 

If  the  first  he  the  same  multiple  of  the  second,  or  the  same  part  of  it,  that  the 
third  is  of  the  fourth;  the  first  is  to  the  second  as  the  third  to  the  fourth. 

First,  if  mA,  mB  be  equimultiples  of  the  magnitudes  A  and  B,  mA  :  A  : : 
mB  :  B. 

Take  of  mA  and  mB  equimultiples  by  any  number  n ;  and  of  A  and  B 
equimultiples  by  any  number  p  ;  these  will  be  nmA  (3.  5.),pA,nmB  (3.  5.), 
pB.  Now,  if  nmA  be  greater  than  pA,  nm  is  also  greater  than  p  ;  and  if 
nm  is  greater  than  p,  nmB  is  greater  than  pB,  therefore,  when  nmA  is  great- 
er than  pA,  nmB  is  greater  than  pB.  In  the  same  manner,  if  nmA=pA, 
nmB=pB,  and  if  nmA/_pA,  nmBZpB.  Now,  nmA,  nmB  are  any  equi- 
multiples of  mA  and  mB ;  and  pA,  pB  are  any  equimultiples  of  A  and  B, 
therefore  mA  :  A  : :  mB  :  B  (def.  5.  5.). 

Next,  Let  C  be  the  same  part  of  A  that  D  is  of  B  ;  then  A  is  tKe  same 
multiple  of  C  that  B  is  of  D,  and  therefore,  as  has  been  demonstrated,  A  ; 
C  : :  B  :  D  and  inversely  (A.  5.)  C  :  A  : :  D  :  B. 

PROP.  C.     THEOR. 

If  the  first  be  to  the  second  as  the  third  to  the  fourth;  and  if  the  first  be  a 
multiple  or  a  part  of  the  second,  the  third  is  the  same  multiple  or  the  same 
part  of  the  fourth. 

Let  A  :  B  : :  0  :  D,  and  first,  let  A  be  a  multiple  of  B,  C  is  the  same 
multiple  of  D,  that  is,  if  A=mB,  C=mD. 

Take  of  A  and  C  equimultiples  by  any  number  as  2,  viz.  2A  and  20  ; 
and  of  B  and  D,  take  equimultiples  by  the  number  2m,  viz.  2mB,  2mD  (3. 


OF  GEOMETRY.    BOOK  V.  113 

5.) ;  then,  because  A=7nB,2A=2mB  ;  and  since  A  :  B  : :  C  :  D,  and  since 
2A=2»iB,  therefore  2C=2mD  (def.  5.  5.),  and  C=wD,  that  is,  C  contains 
D,  m  times,  or  as  often  as  A  contains  B. 

Next,  Let  A  be  a  part  of  B,  C  is  the  same  part  of  D.  For,  since  A  :  B 
•  :  C  :  D,  inversely  (A.  5.),  B  :  A  : :  D  :  C.  But  A  being  a  part  of  B,  B  is 
a  multiple  of  A  ;  and  therefore,  as  is  shewn  above,  D  is  the  same  multiple 
of  C,  and  therefore  C  is  the  same  part  of  D  that  A  is  of  B. 

PROP.  VII.     THEOR. 

Equal  magnitudes  have  the  same  ratio  to  the  same  magnitude  ;  and  the  same 
has  the  same  ratio  to  equal  magnitudes. 

Let  A  and  B  be  equal  magnitudes,  and  C  any  other ;  A  :  C  :  :  B  :  C. 

Let  mA,  wiB,  be  any  equimultiples  of  A  and  B  ;  and  nC  any  multiple 
ofC. 

Because  A=:B,  mA=mB  (Ax.  1.5.);  wherefore,  if  mk  be  greater  than 
TiC,  mB  is  greater  than  nC  ;  and  if  mA=7zC,  »iB=nC  ;  or,  if  mA^nC,  f7*B 
/  wC.  But  mA  and  mB  are  any  equimultiples  of  A  and  B,  and  nC  is  any 
multiple  of  C,  therefore  (def.  5.  5.)  A  :  C  : :  B  ;  C. 

Again,  if  A=B,  C  :  A  :  :  C  :  B  ;  for,  as  has  been  proved,  A  :  C  :  :  B : 
C,  and  inversely  (A.  5.),  C  :  A  :  :  C  :  B. 

PROP.  VIII.     THEOR. 

Of  unequal  magnitudes,  the  greater  has  a  greater  ratio  to  the  same  than  the  less 
has ;  and  the  same  magnitude  has  a  greater  ratio  to  the  less  than  it  has  to 
the  greater. 

Let  A  4-  B  be  a  magnitude  greater  than  A,  and  C  a  third  magnitude, 
A+B  has  to  C  a  greater  ratio  than  A  has  to  0  ;  and  C  has  a  greater  ratio 
to  A  than  it  has  to  A-j-B. 

Let  m  be  such  a  number  that  mA  and  mB  are  each  of  them  greater  than 
C  ;  and  let  nG  be  the  least  multiple  of  C  that  exceeds  mk-\-mQ  ;  then  nC 
—  C,  that  is  (ti— 1)C  (1,  5.)  will  be  less  than  mA+w^B,  or  mA-f-mB,  that 
is,  m(A4-B)  is  greater  than  (n— 1)C.  But  because  nC  is  greater  than 
mA-f-^B,  and  C  less  than  mB,  nC  —  C  is  greater  than  mA,  or  mA  is  less 
than  wC  —  C,  that  is,  than  (n— 1)C.  Therefore  the  multiple  of  A+B  by 
m  exceeds  the  multiple  of  C  by  n— 1,  but  the  multiple  of  A  by  m  does  not 
exceed  the  multiple  of  C  by  n— 1  ;  therefore  i^B  has  a  greater  ratio  to 
C  than  A  has  to  C  (def.  7.  5.). 

Again,  because  the  multiple  of  C  by  n— -1,  exceeds  the  multiple  of  A  by 
m,  but  does  not  exceed  the  multiple  of  A+B  by  m,  C  has  a  greater  ratio  to 
A  than  it  has  to  A-f  B  (def.  7.  5.). 

15 


114  ELEMENTS 


PROP.  IX.     TIIEOR. 

Magnitudes  which  have  the  same  ratio  to  the  same  magnitude  are  equal  to  one 
another  ;  and  those  to  which  the  same  magnitude  has  the  same  ratio  are  equal 
to  one  another. 

If  A:  C::B  :  C,  A=B. 

For  if  not,  let  A  be  greater  than  B  ;  then  because  A  js  greater  than  B, 
two  numbers,  rn  and  w,  may  be  found,  as  in  the  last  proposition,  such  that 
mk  shall  exceed  nC,  while  mB  does  not  exceed  nC.  But  because  A  :  C 
:  :  B  :  C  ;  and  if  wiA  exceed  7iC,  wiB  must  also  exceed  nC  (def.  5.  5.) :  and 
it  is  also  shewn  that  mB  does  not  exceed  7iC,  which  is  impossible.  There- 
fore A  is  not  greater  than  B  ;  and  in  the  same  way  it  is  demonstrated  that 
B  is  not  greater  than  A  ;  therefore  A  is  equal  to  B. 

Next,  let  C  :  A  : :  C  :  B,  A=B.  For  by  inversion  (A.  5.)  A  :  C  :  :  B  : 
C  ;  and  therefore,  by  the  first  case,  A=B. 

PROP.  X.     THEOR. 

That  magnitude,  which  has  a  greater  ratio  than  another  has  to  the  same  magm- 
tudCf  is  the  greatest  of  the  two :  And  that  magnitude,  to  which  the  same  has 
a  greater  ratio  than  it  has  to  another  magnitude ,  is  the  least  of  the  two. 

If  the  ratio  of  A  to  C  be  greater  than  that  of  B  to  C,  A  is  greater  than  B. 

Because  A  :  C  T'B  :  C,  two  numbers  m  and  n  may  be  found,  such  that 
mAynC,  and  mB/^nC  (def.  7.  5.).  Therefore  also  mA'/mB,  and  AT'B 
(Ax.  4.  5.). 

Again,  let  0  :  B/C  :  A;  B^l^A.  For  two  numbers,  ttj  andn  maybe 
found,  such  that  mC/nB,  and  mO/_7iA  (def.  7.  5.).  Therefore,  since  wB 
is  less,  and  nA  greater  than  the  same  magnitude  mC,  nB  Z.  nA,  and  there- 
fore B/ A. 

PROP.  XI.     THEOR. 

Ratios  tnat  are  equal  to  the  same  ratio  are  equal  to  one  another. 

If  A  :  B  :  :  C  :  D ;  and  also  C  :  D  ::  E  :  F  ;  then  A  :  B  :  :  E  :  F. 

Take  mA,  mC,  mE,  any  equimultiples  of  A,  C,  and  E  ;  and  nB,  nD,  wF, 
any  equimultiples  of  B,  D,  and  F.  Because  A  :  B  : :  C  :  D,  if  twA/wB, 
mC/nD  (def.  5.  5.);  birtif  TwC/nD,  mE/nF  (def.  5.  5.),  because  C  :  D 
:  :  E  :  F  ;  therefore  if  mAynB,  mE  ynY.  In  the  same  manner,  if  mA=z 
nB,mE=nF;  and  if  twA/wB,  mE/nF.  Now,  mA,  mE  are  any  equi- 
multiples whatever  of  A  and  E  ;  and  nB,  nF  any  whatever  of  B  and  F  ; 
therefore  A  :  B  :  :  E  ;  F  (def.  5.  5.). 


OF  GEOMETRY.    BOOK  V.  115 

PROP.  XII.    THEOR. 

Jf  any  number  of  magnitudes  be  proportionals,  as  one  of  the  antecedents  is  to 
its  consequent,  so  are  all  the  antecedents,  taken  together,  to  all  the  conse- 
quents. '^ 

If  A  :  B  :  :  C  :  D,  and  C  :  D  :  :  E  :  F ;  then  also,  A  :  B  : :  A+C+E  : 
B+D+F. 

Take  mh.,  mC,  mE  any  equimultiples  of  A,  C,  and  E  ;  and  nB,  nD,  nF, 
any  equimultiples  of  B,  I),  and  F.  Then,  because  A  :  B  : :  C  :  D,  if  mA 
y  wBjmC/'nD  (def.  5.  5.) ;  and  when  mC/nD,  mE/nF,  because  C  :  D 
::  E  :  F.  Therefore,  if  mA/nBjmA+wC+mE/wB+nD+nF  :  In  the 
same,  manner,  if  mx\=/iB,  mA+mC-f-^E=nB+wD+7iF  ;  and  if  mk/_ 
nB,  mA+mC  +  mE/nB  +  nD  +  wF.  Now,  mA+mC+77iE=7w(A+C+ 
E)  (Cor.  1.  5.),  so  that  mA  and  mA+mC+^E  are  any  equimultiples  of 
A,  and  of  A-j-C  +  E.  And  for  the  same  reason  nB,  and  nB+nD+nF  are 
any  equimultiples  of  B,  and  of  B+D+F ;  therefore  (def.  5.  5.)  A  :  B  : : 
A+C+E:  B  +  D-f-F. 

PROP.  XIII.     THEOR. 

If  the  first  have  to  the  second  the  same  ratio  which  the  third  has  to  the  fourth, 
but  the  third  to  the  fourth  a  greater  ratio  than  the  fifth  has  to  the  sixth  ; 
the  first  has  also  to  the  second  a  greater  ratio  than  the  fifth  has  to  the  sixth. 

If  A  :  B  : :  0  :  D  ;  but  C  :  D/E  :  F;  then  also,  A  :  B/E  :  F. 

Because  C  :  D  /E  :  F,  there  are  two  numbers  m  and  n,  such  that  mC  7 
nD,  but  wE  /  wF  (def.  7.  5.).  Now,  if  wC/nD,  mA/nB,  because  A  :  B 
: :  C  :  D.  Therefore  mAjnB,  and  mE/^nF,  wherefore,  A  :  B/E  :  F 
(def.  7.  5.). 

PROP.  XIV.     THEOR. 

If  the  first  have  to  the  seccnd  the  same  ratio  which  the  third  has  to  thefourthy 
and  if  the  first  be  greater  than  the  third,  the  second  shall  be  greater  than 
the  fourth;  if  equal,  equal;  and  if  less,  less. 

If  A:  B::  C:  D;  then  if  A/C,  B/D  ;  if  A=C,B=D;  and  if  A/ 
C,  B/D. 

First,  let  A/C  ;  then  A  :  B/C  :  B  (8.  5.),  but  A  :  B  :  :  C  :  D,  there- 
fore C  :  D/C  :  B  (13.  5.),  and  therefore  B/D  (10.  5.). 

In  the  same  manner,  it  is  proved,  that  if  A=C,  B=D  ;  and  if  A/C, 
BZD. 

PROP.  XV.     THEOR. 
Magnitudes  have  the  same  ratio  to  one  another  which  their  equimultiples  have. 

If  A  and  B  be  two  magnitudes,  and  m  any  number,  A  :  B  . :  mA  :  mB. 
Because  A  :  B  : :  A  :  B  (7.  5.) ;  A  :  B  :  :  A+A  :  B+B  (12.  5.),  or  A  : 


116  ELEMENTS 

• 
B  : :  2A  :  2B.     And  in  the  same  manner,  since  A  :  B  : :  2A  :  2B,  A  :  B 
;  ;  A-f-2A  :  B+2B  (12.  5.),  or  A  :  B  : :  3A  :  3B ;  and  so  on,  for  all  the 
equimultiples  of  A  and  B. 

PROP.  XVI.    THEOR. 

If  four  magnitudes  of  the  same  kind  he  proportionals,  they  will  also  he  pro- 
portionals when  taken  alternately. 

If  A  :  B  : :  C  :  D,  then  alternately,  A  :  C  . :  B  :  D. 

Take  mA,  wiB  any  equimultiples  of  A  and  B,  and  nC,  «D  any  equimul 
tiples  of  C  and  D.  Then  (15.  5.)  A  :  B  : :  mA  :  mB  ;  now  A  :  B  : :  C  : 
D,  therefore  (11.  5.)  C  :  D  : :  mA  :  twB.  ButC  :  D  : :  nC  :  nD  (15.  5.) ; 
therefore  mk  :  mB  :  :  nC  :  nD  (11.  5.) :  wherefore  if  mh-ynG,  mB/nD 
(14.  5.) ;  if  mK=.nC,  ?7iB=wD,  or  if  ttjA^^wC,  wiB^^wD  ;  therefore  (def. 
5.  5.)  A  :  C  : :  B  :  D. 

PROP.  XVII.     THEOR. 

If  magnitudes,  taken  jointly,  he  proportionals,  they  will  also  he  proportionals 
when  taken  separately ;  that  is,  if  the  first,  together  with  the  second,  have 
to  the  second  the  same  ratio  which  the  third,  together  with  the  fourth,  has  to 
the  fourth,  the  first  will  have  to  the  second  the  same  ratio  which  the  third 
has  to  the  fourth. 

If  A+B  :  B  : :  C+D  :  D,  then  by  division  A  :  B  : :  C  :  D. 

Take  mA  and  nB  any  multiples  of  A  and  B,  by  the  numbers  m  and  n ; 
and  first,  let  mA  ynB  :  to  each  of  them  add  mB,  then  mA-^-mB  ymB-\-nB. 
But  mA+mB=m(A+B)  (Cor.  1.  5.),  andmB+nB=(m+n)B  (2.  Cor.  2. 
5.),  therefore  m(A+B)7(m+n)B. 

And  because  A  +  B  :  B  ::  C+D  :  D,  if  m(A+B)7(OT4-n)B,7w(C-f  D) 
7(m+7i)D,  or  mC-{-mT>'/mD-{-nB,  that  is,  taking  ml)  from  both,  mC/ 
nT>.  Therefore,  when  mA  is  greater  than  nB,  mC  is  greater  than  nD.  In 
like  manner  it  is  demonstrated,  that  if  mA=nB,mC=nD,  and  if  »iA/nS, 
that  mDZnJ);  therefore  A  :  B  : :  C  :  D  (def.  5.  5.). 

PROP.  XVIII.    THEOR. 

If  magnitudes,  taken  separately,  he  proportionals,  they  will  also  he  proportion- 
als when  taken  jointly,  that  is,  if  the  first  he  to  the  second  as  the  third  to  the 
fourth,  the  first  and  second  together  will  he  to  the  second  as  the  third  and 
fourth  together  to  the  fourth. 

If  A  :  B  : :  C  :  D,  then,  by  composition,  A+B  :  B  ; :  C+D  :  D. 

Take  m(A+B),  and  tzB  any  multiples  whatever  of  A+B  and  B;  and 
first,  let  m  be  greater  than  n.  Then,  because  A+B  is  also  greater  than 
B,  m(A+B)7nB.  For  the  same  reason,  ?n( C  +  D)/ nD.  In  this  case, 
therefore,  that  is,  when  myn,  7«(A+B)  is  greater  than  nB,  and  ot(C+D) 
is  greater  than  nD.  And  in  the  same  manner  it  may  be  proved,  that  when 
m=n,  m(A+B)  is  greater  than  nB,  and  ?w(C+D)  greater  than  nD. 


OF  GEOMETRY.    BOOK  V.  117 

Next,  let  m/n,  or  n/m,  then  m(A-l-B)  may  be  greater  than  wB,  or  may- 
be equal  to  it,  or  may  be  less  ;  first,  let  W2(A+B)  be  greater  than  nB  ;  then 
also,  wA+mB/nB  ;  take  wiB,  which  is  less  than  nB,  from  both,  and  mA. 
7nB— wBjOr  mAy(n—m)B  {6.  5.).  But  if  7?iA7(n— w)B,  mC7(7i— w) 
D,  because  A  :  B  :  :  C  :  D.  Now,  (n—m)D=nD—mD  (6.  5.),  therefore 
mC/nD— mD,  and  adding  ml)  to  both,  mC+mD/nD,  that  is  (1.  5.), 
»i(C+D)7nD.     If,  therefore,  m(A+B)7nB,m{C-{-D)7nD. 

In  the  same  manner  it  will  be  proved,  that  if  m(A-f- B)=nB,  m(C -{-!)) 
=nD;  and  if  m(A+B)ZnB,  m(C-\''D)ZnT)  ;  therefore  (def.  5.  5.),  A-j- 
B  :  B  ::  C+D  :  D 

PROP.  XIX.    THEOR. 

If  a  whole  magnitude  be  to  a  whole,  as  a  magnitude  taken  from  thefrst  is  to  a 
magnitude  taken  from  the  other ;  the  remainder  will  he  to  the  remainder  as 
the  whole  to  the  whole. 

If  A;  B::  C:  D,  and  if  0  be  less  than  A,  A- C  :  B-D  :  :  A  :  B. 

Because  A  :  B  : :  C  :  D,  alternately  (16. 5.),  A  :  C  :  :  B:  D  ;  and  there- 
fore by  division  (17.  5.)  A— 0  :  C  :  :  B— D  ;  D.  Wherefore,  again  alter- 
nately, A-C  :  B— D  : :  C  :  D  ;  but  A  :  B  :  :  C  :  D,  therefore  (11.  5.)  A 
-C:B-D::A:D. 

Cor.     A-C  :  B-D  :  :  C  :  D. 


PROP.  D.     THEOR. 

If  four  magnitudes  be  proportionals,  they  are  also  proportionals  by  conversion, 
tf^at  is,  the  first  is  to  its  excess  above  the  second,  as  the  third  to  its  excess 
above  the  fourth. 

If  A  :  B  :  :  0  :  D,  by  conversion,  A  :  A— B  :  :  C  :  C— D. 

For,  since  A  :  B  :  :  C  :  D,  by  division  (17.  5.),  A— B  :  B  :  :  C—D  :  D, 
and  inversely  (A.  5.)  B  :  A— B  ;  :  D  :  0— D  ;  therefore,  by  composition 
(18. 5.),  A  :  A-B  ::  C:  C-D. 

CoR.    In  the  same  way,  it  may  be  proved  that  A  :  A+B  :  :  C  :  C  +  D. 
PROP.  XX.     THEOR. 

If  there  be  three  magnitudes,  and  other  three,  which  taken  two  and  two,  have 
the  same  ratio  ;  if  the  first  be  greater  than  the  third,  the  fourth  is  greater 
than  the  sixth ;  if  equal,  equal ;  and  if  less,  less. 

If  there  be  three  magnitudes,  A,  B,  and  C,  and  other  three  D,  E,  and  F ; 
and  if  A  :  B  :  :  D  :  E  ;  and  also  B  :  C  :  :  E  :  F,  then 
if  A/CD/F;  if  A=C,  D=F;  and  if  A^CD 
ZF. 


A,     B,     C, 
D,    E,     F. 


First,  let  A/C;  then  A  :  B/C  :  B  (8.  5.).  But  A  :  B  :  :  D  :  E,  there- 
fore  also  D  :  E/C  :  E  (13. 5.).     Now  B  :  C  :  :  E  :  F,  and  inversely  (A. 


A,     B,       C, 
D,     E,       F. 


118  ELEMENTS 

5.),  C  :  B  :  :  F  :  E  ;  and  it  has  been  shewn  that  D  :  E/C  :  B,  therefore 
D  :  E/F:  E  (13.  5.),  and  consequently  D/F  (10.  5.). 

Next,letA=C;  then  A:  B  ::  C:  B(7.  5.),but  A  :  B  : :  D  :  E  ;  there- 
fore, C  :  B  : :  D  :  E,  but  C  :  B  :  :  F  :  E,  therefore,  D  :  E  :  :  F  :  E  (11. 
5.),  and  D=F  (9.  5.).  Lastly,  let  A/C.  Then  C/ A,  and  because,  as 
was  already  shewn,  C  :  B  : :  F  :  E,  and  B  :  A  :  :  E  :  D  ;  therefore,  by  the 
first  case,  if  C 7 A,  F/D,  that  is,if  A/C,  D/F. 

PROP.  XXL    THEOR. 

If  there  he  three  magnitudes,  and  other  three^  which  have  the  same  ratio  taken  two 
and  two,  hut  in  a  cross  order;  if  the  first  magnitude  he  greater  than  the  third  j 
the  fourth  is  greater  than  the  sixth;  if  equal,  equal ;  and  if  less,  less. 

If  there  be  three  magnitudes.  A,  B,  C,  and  other  three,  D,  E,  and  F, 
such  that  A  :  B  ::  E:  F,andB:  C  ::  D  :  E;  ifA7C,D7F;  if  A=C, 
D=F;  and  if  A/C,  D/F. 

First,  let  A  7  C.  Then  A  :  B  7  C  :  B  (8.  5.),  but 
A:B  ::  E  :  F,  therefore  E  :  F7C  :  B  (13.5.).  Now, 
B  :  C  : :  D  :  E,  and  inversely,  C  :  B  :  :  E  :  D  ;  there- 
fore,E  :  F7E  :  D  (13. 5.),  wherefore,  D 7F  (10.  5.). 

Next,  let  A=C.  Then  (7.  5.)  A  :  B  :  :  C  :  B  ;  but  A  :  B  :  :  E  :  F, 
therefore,  C  :  B  : :  E  :  F  (11.  5.) ;  but  B  :  C  :  :  D  :  E,  and  inversely,  C : 
B  :  :  E  :  D,  therefore  (11.  5.),  E  :  F  : :  E  :  D,  and,  consequently,  D=F 
(9.5.).' 

Lastly,  let  A/C.  Then  C7A,  and,  as  was  already  proved,  C  :  B  : ; 
E  :  D  ;  and  B  :  A  :  :  F  :  E,  therefore,  by  this  first  case,  since  C7A,  F  V 

D,  that  is,  D/F. 

PROP.  XXIL    THEOR. 

If  there  he  any  number  of  magnitudes,  and  as  many  others,  which,  taken  two  ana 
two  in  order,  have  the  same  ratio ;  the  first  will  have  to  the  last  of  the  first 
magnitudes,  the  same  ratio  which  the  first  of  the  other  has  to  the  last* 

First,  let  there  be  three  magnitudes.  A,  B,  C,  and  other  three,  D,  E,  F, 
which,  taken  two  and  two,  in  order,  have  the  same  ratio,  viz.  A  :  B  ;  :  D  : 

E,  and  B  :  C  :  :  E  :  F ;  then  A  :  C  :  :  D  :  F. 

Take  of  A  and  D  any  equimultiples  whatever,  mk,  mD  ;  and  of  B  and 
D  any  whatever,  nB,  nF  :  and  of  C  and  F  any  whatever,  qC,  qF.  Because 
A  :  B  :  :  D  :  E,  mA  :  nB  : :  mD  :  nE  (4.  5.)  ;  and 
for  the  same  reason,  nB  :  ^C  :  :  nE  :  ^F.  Therefore 
(20.  5.)  according  as  mA  is  greater  than  qC,  equal  to 
it,  or  less,  mD  is  greater  than  ^F,  equal  to  it,  or 
less  ;  but  mA,  mD  are  any  equimultiples  of  A  and  D ; 
and  ^C,  ^F  are  any  equimultiples  of  C  and  F ;  therefore  (def  5.  5.),  A  :  C 
: :  D  :  F. 

Again,  let  there  be   four  magnitudes,  and  other  four  which,  taken  two 

*  N.  B.    Tliis  proposition  is  usually  cited  by  the  words  "  ex  8equali,"or  "  ex  aequo." 


A, 

B,     C, 

D, 

E,    F, 

mA, 

nB,  qC, 

mD, 

nE,  qF. 

OF  GEOMETRY.     BOOK  V. 


119 


and  two  in  order,  have  the  same  ratio,  viz.  A:B::  E:F;  B:C::F: 
G ;  C  :  D  : :  G  :  H,  then  A  :  D  : :  E  :  H. 

For,  since  A,  B,  C  are  three  magnitudes,  and 
E,  F,  G  other  three,  vs^hich,  taken  two  and  two, 
have  the  same  ratio,  by  the  foregoing  case,  A  : 
C  :  :  E  :  G.  And  because  also  C  :  D  : :  G  :  H,  by  that  same  case,  A  :  D 
: :  E  :  H.  In  the  same  manner  is  the  demonstration  extended  to  any  num- 
ber of  magnitudes. 


A, 
E, 


B, 
F, 


C, 
G, 


D, 
H. 


PROP.  XXIII.    THEOR. 

If  there  he  any  number  of  magnitudes,  and  as  many  others,  which,  taken  two 
and  two,  in  a  cross  order,  have  the  same  ratio ;  the  first  will  have  to  the  last 
of  the  first  magnitudes  the  same  ratio  which  the  first  of  the  others  has  to 
the  last* 

First,  Let  there  be  three  magnitudes,  A,  B,  C,  and  other  three,  D,  E,  and 
F,  which,  taken  two  and  two  in  a  cross  order,  have  the  same  ratio,  viz.  A 
:  B  : :  E  :  F,  and  B  :  C  : :  D  :  E,  then  A  :  C  : :  D  :  F.  Take  of  A,  B, 
and  D,  any  equimultiples  mk,  mB,  mD  ;  and  of  C,  E,  F  any  equimultiples 
nC,  nE,  nF. 

Because  A  :  B  : :  E  :  F,  and  because  also  A  :  B  : :  mk  :  mB  (15.  5.), 
and  E  :  F  : :  nE  :  TiF  ;  therefore,  mA  :  mB  ::  nE  :  nF  (II.  5.).  Again, 
because  B  :  C  : ;  D  :  E,  mB  :  nC  : :  mD  :  nE  (4. 
5.) ;  and  it  has  been  just  shewn  that  mA  :  mB  : : 
nE  :  nF;  therefore,  if  mA7nC,mD7nF  (21.  5.)  ; 
if  mA=nC,  ?nD=nF;  and  if  mA/wC,  mD/nF. 
Now,  mA  and  mD  are  any  equimultiples  of  A  and 
D,  and  nC,  nF  any  equimultiples  of  C  and  F ;  therefore,  A  :  C 
(def.  5.  5.). 

Next,  Let  theYe  be  four  magnitudes.  A,  B,  0,  and  D,  and  other  four,  E, 
F,  G,  and  H,  which,  taken  two  and  two  in  a  cross  order,  have  the  same 
ratio,  viz.  A  :  B  : :  G  :  H ;  B  :  C  : :  F  :  G,  and 
0  :  D  : :  E  :  F,  then,  A  :  D  : :  E  :  H.  For,since 
A,  B,  C,  are  three  magnitudes,  and  F,  G,  H,  other 
three,  which,  taken  two  and  two,  in  a  cross  order, 
have  the  same  ratio,  by  the  first  case,  A  :  C  : :  F  :  H.  But  C  :  D  : :  E  : 
F,  therefore,  again,  by  the  first  case,  A  :  D  :  :  E  :  H.  In  the  same  manner 
may  the  demonstration  be  extended  to  any  number  of  magnitudes. 


A, 

B, 

c, 

D, 

E, 

F, 

mA, 

mB, 

nC, 

mD, 

nB, 

nF. 

D  :  F 


A, 
E, 


B, 

F, 


C, 
G, 


PROP.  XXIV.    THEOR. 

If  the  first  has  to  the  second  the  same  ratio  which  the  third  has  to  the  fourth ; 
and  the  fifth  to  the  second,  the  same  ratio  which  the  sixth  has  to  the  fourth  ; 
the  first  and  fifth  together,  shall  have  to  the  second,  the  same  ratio  which 
the  third  and  sixth  together,  have  to  the  fourth. 

Let  A  :  B  : :  C  :  D,  and  also  E  :  B  : :  F  :  D,  then  A+E  :  B  : :  C+F  :  D. 

*  N.  B.     This  proposition  is  usually  cited  by  the  words  "  ex  aequali  in  proportione  pertur 
bata ;"  or,  "  ex  aequo  inversely." 


120  ELEMENTS,  &c. 

Because  E  :  B  : :  F  :  D,  by  inversion,  B  :  E  : :  D  :  F.  But  by  hypo- 
thesis, A  :  B  :  :  C  :  D,  therefore,  ex  agquali  (22.  5.),  A  :  E  : :  C  :  F ;  and 
by  composition  (18.  5.),A4-E  :  E  ::  C-j-F:F.  And  again  by  hypothe- 
sis, E  :  B  :  :  F  :  D,  therefore,  ex  aequali  (22.  5.),  A+E  :  B  : :  C-f  F  :  D. 

PROP.  E.    THEOR. 

If  four  magnitudes  he  proportionals,  the  sum  of  the  first  two  is  to  their  diffe- 
rence  as  the  sum  of  the  other  two  to  their  difference. 

Let  A  :  B  : :  C  :  D  ;  then  if  A/B, 

A-f-B  :  A-B  ::  C  +  D  :  C-D;  orifA^B 

A+B  :  B-A  : :  C+D  :  D-C. 
For,  if  A/B,  then  because  A  :  B  : :  C  :  D,  by  division  (17.  5.), 

A— B  :  B  ; :  C— D  :  D,  and  by  inversion  (A.  5.), 

B  :  A— B  :  :  D  :  C— D.     But,  by  composition  (18.  5.), 

A+B  :  B  :  :  C-f-D  :  D,  therefore,  ex  aequali  (22.  5.), 

A+B  :  A~B  ::  C+D  :  C-D. 
In  the  same  manner,  if  B  /  A,  it  is  proved,  that 

A+B:  B-A::  C+D:  D-C. 

PROP.  F.    THEOR. 
Ratios  which  are  compounded  of  equal  ratios,  are  equal  to  one  another. 

Let  the  ratios  of  A  to  B,  and  of  B  to  C,  which  compound  the  ratio  of  A 
to  C,  be  equal,  each  to  each,  to  the  ratios  of  D  to  E,  and  E  to  F,  which  com- 
pound the  ratio  of  D  to  F,  A  :  C  : :  D  :  F. 

For,  first,  if  the  ratio  of  A  to  B  be  equal  to  that  of 
D  to  E,  and  the  ratio  of  B  to  C  equal  to  that  of  E  to 
F,  ex  aequali  (22.  5.),  A  :  C  :  :  D  :  F. 

And  next,  if  the  ratio  of  A  to  B  be  equal  to  that  of  E  to  F,  and  the  ratio 
of  B  to  C  equal  to  that  of  D  to  E,  ex  aequali  inversely  (23.  5.),  A  :  C  : :  D 
:  F.  In  the  same  manner  may  the  proposition  be  demonstrated,  whatever 
be  the  number  of  ratios. 

PROP.  G.    THEOR. 

If  a  magnitude  measure  each  of  two  others,  it  will  also  measure  their  sum  and 

difference. 

Let  C  measure  A,  or  be  contained  in  it  a  certain  number  of  times  ;  9  times 
for  instance  :  let  C  be  also  contained  in  B,  suppose  5  times.  Then  A=9C, 
and  B=5C  ;  consequently  A  and  B  together  must  be  equal  to  14  times  C, 
so  that  C  measures  the  sum  of  A  and  B  ;  likewise,  since  the  difference  of 
A  and  B  is  equal  to  4  times  C,  C  also  measures  this  difference.  And  had 
any  other  numbers  been  chosen,  it  is  plain  that  the  results  would  have  been 
similar.  For,  let  A=mC,  and  B=nC  ;  A+B=:(m+n)C,  and  A— B= 
(m— n)C. 

Cor.  If  C  measure  B,  and  also  A— B,  or  A+B,  t^  must  measure  A,  for 
the  sum  of  B  and  A— B  is  A,  and  the  difference  of  B  and  A+B  is  also  A. 


A,     B,     C, 
D,     E,     F. 


ELEMENTS 


OF 


GEOMETRY 


BOOK  VI. 

DEFINITIONS 


1.  Similar  rectilineal  figures  are 
those  which  have  their  several 
angles  equal,  each  to  each,  and 
the  sides  about  the  equal  angles 
proportionals. 
In  two  similar  figures,  the  sides  which  lie  adjacent  to  equal  angles,  are 

called  homologous  sides.  Those  angles  themselves  are  called  homo- 
logous angles.  In  different  circles,  similar  arcs,  sectors,  and  segments, 
are  those  of  which  the  arcs  subtend  equal  angles  at  the  centre.  Two 
equal  figures  are  always  similar  ;  but  two  similar  figures  may  be  very 
unequal. 

2.  Two  sides  of  one  figure  are  said  to  be  reciprocally  proportional  to  two 
sides  of  another,  when  one  of  the  sides  of  the  first  is  to  one  of  the 
sides  of  the  second,  as  the  remaining  side  of  the  second  is  to  the  re- 
maining side  of  the  first. 

3.  A  straight  line  is  said  to  be  cut  in  extreme  and  mean  ratio,  when  the 
whole  is  to  the  greater  segment,  as  the  greater  segment  is  to  the  less. 

4.  The   altitude  of  a  triangle  is   the  straight  line 

drawn  from  its  vertex  perpendicular  to  the  base. 

The  altitude  of  a  parallelogram  is  the  perpendicu- 
lar which  measures  the  distance  of  two  oppo- 
site sides,  taken  as  bases.  And  the  altitude  of 
a  trapezoid  is  the  perpendicular  drawn  between 
its  two  parallel  sides. 

PROP.  I.     THEOR. 

Triangles  and  parallelograms,  of  the  same  altitude,  are  one  to  another  as  their 

bases. 

Let  the  triangles  ABC,  ACD,  and  the  parallelograms  EC,  CF  have  the 
same  altitude,  viz.  the  perpendicular  drawn  from  the  point  A  to  BD :  Then, 

16 


122  ELEMENTS 

as  the  base  BC,  is  to  the  base  CD,  so  is  the  triangle  ABC  to  the  triangle 
ACD,  and  the  parallelogram  EC  to  the  parallelogram  CF. 

Produce  BD  both  ways  to  the  points  H,  L,  and  take  any  number  of 
straight  lines  BG,  GH,  each  equal  to  the  base  BC;  and  DK,  KL,  any 
number  of  them,  each  equal  to  the  base  CD  ;  and  join  AG,  AH,  AK,  AL. 
Then,  because  CB,  BG,  GH  are  all  equal,  the  triangles  AHG,  AGB,  ABC 
are  all  equal  (38. 1.) ;  Therefore,  whatever  multiple  the  base  HC  is  of  the 
base  BC,  the  same  multiple  is  the  triangle  AHC  of  the  triangle  ABC.  For 
the  same  reason,  whatever  the  base  LC  is  of  the  base  CD,  the  same  mul- 
tiple is  the  triangle  ALC  of 

the  triangle    ADC.      But  if  E     A.         F 

the  base  HC  be  equal  to  the 
base  CL,  the  triangle  AHC 
is  also  equal  to  the  triangle 
ALC  (38.  1.):  and  if  the 
base  HC  be  greater  than  the 
base  CL,  likewise  the  trian- 
gle AHC  is  greater  than  the 
triangle  ALC  ;  and  if  less,  jx  n^  ^ 
less.  Therefore,  since  there 
are  four  magnitudes,  viz.  the  two  bases  BC,  CD,  and  the  two  triangles 
ABC,  ACD  ;  and  of  the  base  BC  and  the  triangle  ABC,  the  first  and  third, 
any  equimultiples  whatever  have  been  taken,  viz.  the  base  HC,  and  the 
triangle  AHC  ;  and  of  the  base  CD  and  triangle  ACD,  the  second  and 
fourth,  have  been  taken  any  equimultiples  whatever,  viz.  the  base  CL  and 
triangle  ALC  ;  and  since  it  has  been  shewn,  that  if  the  base  HC  be  greater 
than  the  base  CL,  the  triangle  AHC  is  greater  than  the  triangle  ALC  ; 
and  if  equal,  equal ;  and  if  less,  less  ;  Therefore  (def.  5.  5.),  as  the  base 
BC  is  to  the  base  CD,  so  is  the  triangle  ABC  to  the  triangle  ACD. 

And  because  the  parallelogram  CE  is  double  of  the  triangle  ABC  (41. 
1.),  and  the  parallelogram  CF  double  of  the  triangle  ACD,  and  because 
magnitudes  have  the'  same  ratio  which  their  equimultiples  have  (15.  5.) ; 
as  the  triangle  ABC  is  to  the  triangle  ACD,  so  is  the  parallelogram  EC  to 
the  parallelogram  CF.  And  because  it  has  been  shewn,  that,  as  the  base 
BC  is  to  the  base  CD,  so  is  the  triangle  ABC  to  the  triangle  ACD  ;  and 
as  the  triangle  ABC  to  the  triangle  ACD,  so  is  the  parallelogram  EC  to 
the  parallelogram  CF  ;  therefore,  as  the  base  BC  is  to  the  base  CD,  so  is 
(11.  5.)  the  parallelogram  EC  to  the  parallelogram  CF. 

CoR.  From  this  it  is  plain,  that  triangles  and  parallelograms  that  have 
equal  altitudes,  are  to  one  another  as  their  bases. 

Let  the  figures  be  placed  so  as  to  have  their  bases  in  the  same  straight 
line  ;  and  having  drawn  perpendiculars  from  the  vertices  of  the  triangles  to 
the  bases,  the  straight  line  which  joins  the  vertices  is  parallel  to  that  in 
which  their  bases  are  (33.  1.),  because  the  perpendiculars  are  both  equal 
and  parallel  to  one  another.  Then,  if  the  same  construction  be  made  as  in 
the  proposition,  the  demonstration  will  be  the  same. 


OF  GEOMETRY.    BOOK  VI. 


123 


PROP.  II.     THEOR. 

7/  a  straight  line  he  drawn  parallel  to  one  of  the  sides  of  a  triangle,  it  will  cut 
the  other  sides,  or  the  other  sides  produced,  proportionally :  And  if  the 
sides,  or  the  sides  produced,  be  cut  proportionally,  the  straight  line  which 
joins  the  points  of  section  will  be  parallel  to  the  remaining  side  of  the  tri- 
angle. 

Let  DE  be  drawn  parallel  to  BC,  one  of  the  sides  of  tlie  triangle  ABC 
BD  is  to  DA  as  CE  to  EA. 

Join  BE,  CD  ;  then  the  triangle  BDE  is  equal  to  the  triangle  CDE  (37 
1.),  because  they  are  on  the  same  base  DE  and  between  the  same  paral 
lels  DE,  BC :  but  ADE  is  another  triangle,  and  equal  magnitudes  have 
to  the  same,  the  same  ratio  (7.  5.)  ;  therefore,  as  the  triangle  BDE  to  the 
triangle  ADE,  so  is  the  triangle  CDE  to  the  triangle  ADE  ;  but  as  the 
triangle  BDE  to  the  triangle  ADE,  so  is  (1.  6.)  BD  to  DA,  because,  hav- 
ing the  same  altitude,  viz.  the  perpendicular  drawn  from  the  point  E  to  AB, 
they  are  to  one  another  as  their  bases  ;  and  for  the  same  reason,  as  the 
triangle  CDE  to  the  triangle  ADE,  so  is  CE  to  Bk.  Therefore,  as  BD 
to  DA,  so  is  CE  to  EA  (11.  5.). 

Next,  let  the  sides  AB,  AC  of  the  triangle  ABC,  or  these  sides  produced, 


be  cut  proportionally  in  the  points  D,  E,  that  is,  so  that  BD  be  to  DA,  as 
CE  to  EA,  and  join  DE  ;  DE  is  parallel  to  BC. 

The  same  construction  being  made,  because  as  BD  to  DA,  so  is  CE  to 
EA  ;  and  as  BD  to  DA,  so  is  the  triangle  BDE  to  the  triangle  ADE  (1.6.): 
and  as  CE  to  EA,  so  is  the  triangle  CDE  to  the  triangle  ADE  ;  therefore 
the  triangle  BDE,  is  to  the  triangle  ADE,  as  the  triangle  CDE  to  the  tri- 
angle ADE  ;  that  is,  the  triangles  BDE,  CDE  have  the  same  ratio  to  the 
triangle  ADE  ;  and  therefore  (9.  5.)  the  triangle  BDE  is  equal  to  the  tri- 
angle CDE  :  And  they  are  on  the  same  base  DE  ;  but  equal  triangles  on 
the  same  base  are  between  the  same  parallels  (39.  1.) ;  therefore  DE  is 
parallel  to  BC. 


124 


ELEMENTS 


PROP.  III.     THEOR. 

If  the  angle  of  a  triangle  be  bisected  by  a  straight  line  which  also  cuts  the  base  ; 
the  segments  of  the  base  shall  have  the  same  ratio  which  the  other  sides  of 
the  triangle  have  to  one  another  ;  And  if  the  segments  of  the  base  have  the 
same  ratio  which  the  other  sides  of  the  triangle  have  to  one  another  ^  the  straight 
line  drawn  from  the  vertex  to  the  point  of  section,  bisects  the  vertical  angle. 

Let  the  angle  BAG,  of  any  triangle  ABC,  be  divided  into  two  equal  an- 
gles, by  the  straight  line  AD  ;  BD  is  to  DC  as  BAto  AC. 

Through  the  point  C  draw  CE  parallel  (Prop.  31.  1.)  to  DA,  and  let  BA 
produced  meet  CE  inE.  Because  the  straight  line  AC  meets  the  paral- 
lels AD,  EC,  the  angle  ACE  is  equal  to  the  alternate  angle  CAD  (29.  L) : 
But  CAD,  by  the  hypothesis,  is  equal  to  the  angle  BAD  ;  wherefore  BAD 
is  equal  to  the  angle  i^E.  Again, 
because  the  straight  line  BAE  meets 
the  parallels  AD,  EC, the  exterior  an- 
gle BAD  is  equal  to  the  interior  and  i 
opposite  angle  AEC  ;  But  the  angle 
ACE  has  been  proved  equal  to  the  an- 
gle BAD  ;  therefore  also  ACE  is 
equal  to  the  angle  AEC,  and  conse- 
quently the  side  AE  is  equal  to  the 
side  (6.  1 .)  AC.  And  because  AD  is 
drawn  parallel  to  one  of  the  sides  of 
the  triangle  BCE,  viz.  to  EC,  BD  is 
to  DC,  as  BA  to  AE  (2.  6.) ;  but  AE  is  equal  to  AC  ;  therefore,  as  BD  to 
DC,  so  is  BAto  AC  (7.  5.). 

Next,  let  BD  be  to  DC,  as  BA  to  AC,  and  join  AD  ;  the  angle  BAC  is 
divided  into  two  equal  angles,  by  the  straight  line  AD. 

The  same  construction  being  made  •  because,  as  BD  to  DC,  so  is  BA 
to  AC ;  and  as  BD  to  DC,  so  is  BA 
to  AE  (2.  6.),  because  AD  is  paral- 
lel to  EC  :  therefore  AB  is  to  AC,  as 
AB  to  AE  (11.  5.)  :  Consequently 
AC  is  equal  to  AE  (9.  5.),  and  the 
angle  AEC  is  therefore  equal  to  the 
angle  ACE  (5.  1.).  But  the  angle 
AEC  is  equal  to  the  exterior  and  op- 
posite angle  BAD  ;  and  the  angle 
ACE  is  equal  to  the  alternate  angle 
CAD  (29.  1.):  Wherefore  also  the 
angle  BAD  is  equal  to  the  angle 
CAD  :  Therefore  the  angle  BAC  is  cut  into  two  equal  angles  by  the  straight 
line  AD. 


OF  GEOMETRY.  BOOK  VI.  l^ 


PROP.  A.  THEOR. 

If  the  exterior  angle  of  a  triangle  he  bisected  by  a  straight  line  which  also  cuts 
the  base  produced ;  the  segments  between  thebisecting  line  and  the  extremities 
of  the  base  have  the  same  ratio  which  the  other  sides  of  the  triangles  have  to 
one  another  ;  And  if  the  segments  of  the  base  produced  have  the  same  ratio 
which  the  other  sides  of  the  triangles  have,  the  straight  line^  drawn  from  the 
vertex  to  the  point  of  section,  bisects  the  exterior  angle  of  the  triangle. 

Let  the  exterior  angle  CAE,  of  any  triangle  ABC,  be  bisected  by  the 
straight  line  AD  which  meets  the  base  produced  in  D  ;  BD  is  to  DC,  as 
BA  to  AC. 

Through  C  draw  CF  parallel  to  AD  (Prop.  31.  1.) :  and  because  the 
straight  line  AC  meets  the  parallels  AD,  FC,  the  angle  ACF  is  equal  to 
the  alternate  angle  CAD  (29.  1.):  But  CAD  is  equal  to  the  angle  DAE 
(Hyp.)  :  therefore  also  DAE  is  equal  to  the  angle  ACF.  Again,  because 
the  straight  line  FAE  meets  the  parallels  AD,  FC,  the  exterior  angle  DAE 
is  equal  to  the  interior  and  opposite  angle  CFA ;  But  the  angle  ACF  has 
been  proved  to  be  equal  to  the  an- 
gle DAE  ;  therefore  also  the  angle 
ACF  is  equal  to  the  angle  CFA, 
and  consequently  the  side  AF  is 
equal  to  the  side  AC  (6.  1.) ;  and, 
because  AD  is  parallel  to  FC,  a 
side  of  the  triangle  BCF,  BD  is  to 
DC,  as  BA  to  AF  (2.  6.) ;  but  AF 
is  equal  to  AC ;  therefore  as  BD 
is  to  DC,  so  is  BA  to  AC. 

Now  let  BD  be  to  DC,  as  BA  to  AC,  and  join  AD  ;  the  angle  CAD  is 
equal  to  the  angle  DAE. 

The  same  construction  being  made,  because  BD  is  to  DC  as  BA  to  AC  ; 
and  also  BD  to  DC,  BA  to  AF  (2.  6. ) ;  therefore  BA  is  to  AC,  as  BA  to 
AF  (11.  5.),  wherefore  AC  is  equal  to  AF  (9.  5.),  and  the  angle  AFC 
equal  (5.  1.)  to  the  angle  ACF  :  but  the  angle  AFC  is  equal  to  the  exte- 
rior angle  EAD,  and  the  angle  ACF  to  the  alternate  angle  CAD  ;  there- 
fore also  EAD  is  equal  to  the  angle  CAD 

PROP.  IV.     THEOR. 

The  sides  about  the  equal  angles  of  equiangular  triangles  are  proportionals ;  and 
those  which  are  opposite  to  the  equal  angles  are  homologous  sides,  that  is,  are 
the  antecedents  or  consequents  of  the  ratios 

Let  ABC,  DCE,  be  equiangular  triangles,  having  the  angle  ABC  equal 
to  the  angle  DCE,  and  the  angle  ACB  to  the  angle  DEC,  and  conse- 
quently (4.  Cor.  32.  1.)  the  angle  BAC  equal  to  the  angle  CDE.  The 
sides  about  the  equal  angles  of  the  triangles  ABC,  DCE  are  proportionals , 
and  those  are  the  homologous  sides  which  are  opposite  to  the  equal  an- 
gles. 


1^ 


ELEMENTS 


Let  the  triangle  DCE  be  placed,  so  that  its  side  CE  may  be  contiguous 
to  BC,  and  in  the  same  straight  line  with  it :  And  because  the  angles  ABC, 
ACB  are  together  less  than  two  right  angles  (17.  1.),  ABC  and  DEC, 
which  is  equal  to  ACB,  are  also  less  than 
two  right  angles  :  wherefore  BA,  ED  pro- 
duced shall  meet  (1  Cr.  29.1.);  let  them  be  pro- 
duced and  meet  in  the  point  F  ;  and  because 
the  angle  ABC  is  equal  to  the  angle  DCE, 
BF  is  parallel  (28.  1.)  to  CD.  Again,  be- 
cause the  angle  ACB  is  equal  to  the  angle 
DEC,  AC  is  parallel  to  FE  (28. 1.) :  There- 
fore FACD  is  a  parallelogram  ;  and  conse- 
quently AF  is  equal  to  CD,  and  AC  to  FD 
(34.  1.) :  And  because  AC  is  parallel  to  FE, 
one  of  the  sides  of  the  triangle  FBE,  BA  :  AF  : :  BC  :  CE  (2.  6.) :  but 
AF  is  equal  to  CD  ;  therefore  (7.  5.)  BA  :  CD  : :  BC  :  CE  ;  and  alter- 
nately, BA  :  BC  :  :  DC  :  CE  (16.  5.)  :  Again,  because  CD  is  parallel  to 
BF,  BC  :  CE  : :  FD  :  DE  (2.  6.) ;  but  FD  is  equal  to  AC  ;  therefore  BC 
:  CE  : :  AC  :  DE  ;  and  alternately,  BC  :  CA  : :  CE  :  ED.  Therefore, 
because  it  has  been  proved  that  AB  :  BC  : :  DC  :  CE ;  and  BC  :  CA  ; : 
CE  :  ED,  ex  aequali,  BA  :  AC  ;  :  CD  :  DE. 


PROP.  V.     THEOR. 


If  the  sides  of  two  triangles,  about  each  of  their  angles y  he  proportionals ,  the 
triangles  shall  he  equiangular^  and  have  their  equal  angles  opposite  to  the 
homologous  sides. 

Let  the  triangles  ABC,  DEF  have  their  sides  proportionals,  so  that  AB 
is  to  BC,  as  DE  to  EF  ;  and  BC  to  CA,  as  EF  to  FD  ;  and  consequently 
ex  aequali,  BA  to  AC,  as  ED  to  DF  ;  the  triangle  ABC  is  equiangular  to 
the  triangle  DEF,  and  their  equal  angles  are  opposite  to  the  homologous 
sides,  viz.  the  angle  ABC  being  equal  to  the  angle  DEF,  and  BCA  to 
EFD,  and  also  BAC  to  EDF. 

At  the  points  E,  F,  in  the  straight 
line  EF,  make  (Prop. 23.  l.)the  an- 
gle FEG  equal  to  the  angle  ABC, 
and  the  angle  EFG  equal  to  BCA, 
wherefore  the  remaining  angle  BAC 
is  equal  to  the  remaining  angle 
EGF  (4.  Cor.  32.  1.),  and  the  trian- 
gle ABC  is  therefore  equiangular  to 
the  triangle  GEF  ;  and  consequently 
they  have  their  sides  opposite  to  the 
equal  angles  proportionals  (4.  6.). 
Wherefore, 

AB  :  BC  : :  GE  :  EF ;  but  by  supposition, 

AB  :  BC  : :  DE  :  EF,  therefore, 

DE  :  EF  : :  GE  :  EF.     Therefore  (11.  5.)  DE  and  GE  have 


OF  GEOMETRY.     BOOK  VI.  127 

the  same  ratio  to  EF,  and  consequently  are  equal  (9.  5.).  For  the  same 
reason,  DF  is  equal  to  FG  :  And  because,  in  the  triangles  DEF,  GEF, 
DE  is  equal  to  EG,  and  EF  common,  and  also  the  base  DF  equal  to  the 
base  GF  ;  therefore  the  angle  DEF  is  equal  (8.  l.)to  the  angle  GEF,  and 
the  other  angles  to  the  other  angles,  which  are  subtended  by  the  equal 
sides  (4.  1.).  Wherefore  the  angle  DFE  is  equal  to  the  angle  GFE,  and 
EDF  to  EGF :  and  because  the  angle  DEF  is  equal  to  the  angle  GEF, 
and  GEF  to  the  angle  ABC  ;  therefore  the  angle  ABC  is  equal  to  the  an- 
gle DEF :  For  the  same  reason,  the  angle  ACB  is  equal  to  the  angle 
DFE,  and  the  angle  at  A  to  the  angle  at  D.  Therefore  the  triangle  ABC 
is  equiangular  to  the  triangle  DEF. 


PROP.  VI.     THEOR. 

If  two  triangles  have  one  angle  of  the  one  equal  to  one  angle  of  the  other ^  and 
the  sides  about  the  equal  angles  proportionals,  the  triangles  shall  be  equian- 
gular, and  shall  have  those  angles  equal  which  are  opposite  to  the  homolo' 
gous  sides. 

Let  the  triangles  ABC,  DEF  have  the  angle  BAC  in  the  one  equal  to 
the  angle  EDF  in  the  other,  and  the  sides  about  those  angles  proportion- 
als ;  that  is,  BA  to  AC,  as  ED  to  DF  ;  the  triangles  ABC,  DEF  are  equi- 
angular, and  have  the  angle  ABC  equal  to  the  angle  DEF,  and  ACB  to 
DFE. 

At  the  points  D,  F,  in  the 
straight  line  DF,  make  (Prop. 
23.  1.)  the  angle  FDG  equal  to 
either  of  the  angles  BAC,  EDF  ; 
and  the  angle  DFG  equal  to  the 
angle  ACB  ;  wherefore  the  re- 
maining angle  at  B  is  equal  to 
the  remaining  one  at  G  (4.  Cor. 
32.  1.),  and  consequently  the 
triangle  ABC  is  equiangular  to 
the  triangle  DGF  ;  and  therefore 

BA  :  AC  :  :  GD  (4.  6.)  :  DF.     But  by  hypothesis, 

BA  :  AC  :  :  ED  :  DF ;  and  therefore 

ED  :  DF  : :  GD  :  (11.  5.)  DF  ;  wherefore  ED  is  equal  (9.  5.)  to 
DG ;  and  DF  is  common  to  the  two  triangles  EDF,  GDF  ;  therefore  the 
two  sides  ED,  DF  are  equal  to  the  two  sides  GD,  DF ;  but  the  angle 
EDF  is  also  equal  to  the  angle  GDF  ;  wherefore  the  base  EF  is  equal  to 
the  base  FG  (4.  1.),  and  the  triangle  EDF  to  the  triangle  GDF,  and  the 
remaining  angles  to  the  remaining  angles,  each  to  each,  which  are  sub- 
tended by  the  equal  sides  :  Therefore  the  angle  DFG  is  equal  to  the  angle 
DFE,  and  the  angle  at  G  to  the  angle  at  E  :  But  the  angle  DFG  is  equal 
to  the  angle  ACB  ;  therefore  the  angle  ACB  is  equal  the  angle  DFE,  and 
the  angle  BAC  is  equal  to  the  angle  EDF  (Hyp.) ;  wherefore  also  the  re- 
maining angle  at  B  is  equal  to  the  remaining  angle  at  E.  Therefore  th© 
triangle  ABC  is  equiangular  to  the  triangle  DEF. 


128  ELEMENTS 

PROP.  VII.     THEOR. 

If  two  triangles  have  one  angle  of  the  one  equal  to  one  angle  of  the  other ^  and 
the  sides  about  two  other  angles  proportionals,  then,  if  each  of  the  remaining 
angles  be  either  less,  or  not  less,  than  a  right  angle,  the  triangles  shall  be 
equiangular,  and  have  those  angles  equal  about  which  the  sides  are  propor- 
tionals. 

Let  the  two  triangles  ABC,  DEF  have  one  angle  in  the  one  equal  to  one 
angle  in  the  other,  viz.  the  angle  BAG  to  the  angle  EDF,  and  the  sides 
about  two  other  angles  ABC,  DEF  proportionals,  so  that  AB  is  to  BC,  as 
DE  to  EF  ;  and,  in  the  first  case,  let  each  of  the  remaining  angles  at  C,  F, 
be  less  than  a  right  angle.  The  triangle  ABC  is  equiangular  to  the  tri- 
angle DEF,  that  is,  the  angle  ABC  is  equal  to  the  angle  DEF,  and  the 
remaining  angle  at  C  to  the  remaining  angle  at  F. 

For,  if  the  angles  ABC,  DEF  be  not  equal,  one  of  them  is  greater  than 
the  other  :  Let  ABC  be  the  greater,  and  at  the  point  B,  in  the  straight 
line  x\B,  make  the  angle  ABG  equal 
to  the  angle  (Prop.  23.  l.)DEF  :  and 
because  the  angle  at  A  is  equal  to  the 
angle  at  D,  and  the  angle  ABG  to 
the  angle  DEF ;  the  remaining  an- 
gle AGB  is  equal  (4.  Cor.  32.  1.)  to 
the  remaining  angle  DFE ;  There- 
fore the  triangle  ABG  is  equiangular 
to  the  triangle  DEF ; 

wherefore  (4.  6.),  AB  :  BG  : :  DE  ;  EF ;  but, 
by  hypothesis,       DE  :  EF  : :  AB  :  BC, 
therefore,  AB  :  BC  :  :  AB  :  BG  (11.  5.), 

and  because  AB  has  the  same  ratio  to  each  of  the  lines  BC,  BG ;  BC  is 
equal  (9.  5.)  to  BG,  and  therefore  the  angle  BGC  is  equal  to  the  angle 
BCG  (5.  L) ;  But  the  angle  BCG  is,  by  hypothesis,  less  than  a  right  an- 
gle ;  therefore  also  the  angle  BGC  is  less  than  a  right  angle,  and  the  adja- 
cent angle  AGB  must  be  greater  than  a  right  angle  (13. 1.).  But  it  was 
proved  that  the  angle  AGB  is  equal  to  the  angle  at  F  ;  therefore  the  angle 
at  F  is  greater  than  a  right  angle  :  But  by  the  hypothesis,  it  is  less  than  a 
right  angle  ;  which  is  absurd.  Therefore  the  angles  ABC,  DEF  are  not 
unequal,  that  is,  they  are  equal :  And  the  angle  at  A  is  equal  to  the  angle 
at'D  ;  wherefore  the  remaining  angle  at  C  is  equal  to  the  remaining  angle 
at  F ;  Therefore  the  triangle  ABC  is  equiangular  to  the  triangle  DEF. 

Next,  let  each  of  the  angles  at  C,  F  be  not  less  than  a  right  angle  ;  the 
triangle  ABC  is  also,  in  this  case,  equiangular  to  the  triangle  DEF. 

The  same  construction  being 
made,  it  may  be  proved,  in  like  A. 

manner,  that  BC  is  equal  to  BG,  ^/^  yi 

and  the  angle  at  C  equal  to  the  ^^  /  - 

angle  BGC :  But  the  angle  at  C  y^       I 

is  not  less  than  a  right  angle  ;  xl^S— ""Z  6* 

therefore  the  angle  BGC  is  not  ^    '       — ^       _^ 

less  than  a  right  angle :  Where-      B  C      Ji* 


OF  GEOMETRY.    BOOK  VI. 


129 


fore,  two  angles  of  the  triangle  BGC  are  together  not  less  than  two  right 
angles,  which  is  impossible  (17.  1.) ;  and  therefore  the  triangle  ABC  may- 
be proved  to  be  equiangular  to  the  triangle  DEF,  as  in  the  first  case. 

PROP.  VIII.    THEOR. 

In  a  rigid  angled  triangle  if  a  perpendicular  be  drawn  from  the  right  angle  to 
the  base ;  the  triangles  on  each  side  of  it  are  similar  to  the  whole  triangle, 
and  to  one  another. 

Let  ABC  be  a  right  angled  triangle,  having  the  right  angle  BAC  ;  and 
from  the  point  A  let  AD  be  drawn  perpendicular  to  the  base  BC  :  the  trian- 
gles ABD,  ADC  are  similar  to  the  whole  triangle  ABC,  and  to  one  another. 

Because  the  angle  BAC  is  equal  to  the  angle  ADB,  each  of  them  being 
a  right  angle,  and  the  angle  at  B  com- 
mon to  the  two  triangles  ABC,  ABD ; 
the  remaining  angle  ACB  is  equal  to 
the  remaining  angle  BilD  (4.  Cor.  32. 
1.):  therefore  the  triangle  ABC  is 
equiangular  to  the  triangle  ABD,  and 
the  sides  about  their  equal  angles  are 
proportionals  (4.  6.) ;  wherefore  the 
triangles  are  similar  (def.  1.  6.).  In 
like  manner,  it  may  be  demonstrated,  that  the  triangle  ADC  is  equiangular  and 
similar  to  the  triangle  ABC  :  and  the  triangles  ABD,  ADC,  being  each  equi- 
angular and  similar  to  ABC,  and  equiangular  and  similar  to  one  another. 

CoR.  From  this  it  is  manifest,  that  the  perpendicular,  drawn  from  the 
right  angle  of  a  right  angled  triangle,  to  the  base,  is  a  mean  proportional 
between  the  segments  of  the  base  ;  and  also  that  each  of  the  sides  is  a  mean 
proportional  between  the  base,  and  its  segment  adjacent  to  that  side.  For 
in  the  triangles  BDA,  ADC, 


BD 

DA  : 

:  DA 

DC  (4.  6.)  ;  and  in    the 

triangles    ABC,    BDA,    BC 

:  BA  : 

:  BA  : 

BD  (4.  6.)  ;  and  in    the 

triangles    ABC,    ACD,    BC 

:  CA  : 

:  CA  : 

CD  (4.  6.). 

PROP.  IX.     PROB. 

From  a  given  straight  line  to  cut  off  any  part  required^  that  is,  a  part  which 
shall  be  contained  in  it  a  given  number  of  times. 

Let  AB  be  the  given  straight  line  ;  it  is  required 
to  cut  off  from  AB,  a  part  which  shall  be  contained 
in  it  a  given  number  of  times. 

From  the  point  A  draw  a  straight  line  AC  mak- 
ing any  angle  with  AB  ;  and  in  AC  take  any  point 
D,  and  take  AC  such  that  it  shall  contain  AD,  as 
oft  as  AB  is  to  contain  the  part,  which  is  to  be  cut 
off  from  it ;  join  BC,  and  draw  DE  parallel  to  it: 
then  AE  is  the  part  required  to  be  cut  off. 

Because  ED  is  parallel  to  one  of  the  sides  of  the 

17 


130 


ELEMENTS 


triangle  ABC,  viz.  to  BC,  CD  :  DA  : :  BE  :  EA  (2.  6.) ;  and  by  composi- 
tion (18.  5),  CA  :  AD  : :  BA  :  AE :  But  CA  is  a  multiple  of  AD  ;  there- 
fore (C.  5.)  BA  is  the  same  multiple  of  AE,  or  contains  AE  the  same  num- 
ber of  times  that  AC  contains  AD  ;  and  therefore,  whatever  part  AD  is  of 
AC,  AE  is  the  same  of  AB  ;  wherefore,  from  the  straight  line  AB  the  par* 
required  is  cut  off. 


PROP.  X.     PROB. 

To  divide  a  given  straight  line  similarly  to  a  given  divided  straight  line,  that  w, 
into  parts  that  shall  have  the  same  ratios  to  one  another  which  the  parts  of 
the  divided  given  straight  line  have. 

Let  AB  be  the  straight  line  given  to  be  divided,  and  AC  the  divided  line, 
it  is  required  to  divide  AB  similarly  to  AC. 

Let  AC  be  divided  in  the  points  D,  E  ;  and  let  AB,  AC  be  placed  so  as 
to  contain  any  angle,  and  join  BC,  and  through  the  points  D,  E,  draw 
(Prop.  31.  1.)  DF,  EG,  parallel  to  BC  ;  and 
through  D  draw  DHK,  parallel  to  AB  ;  there- 
fore each  of  the  figures  FH,  HB,  is  a  parallelo- 
gram :  wherefore  DH  is  equal  (34. 1.)  to  FG, 
and  HK  to  GB  :  and  because  HE  is  parallel 
to  KC,  one  of  the  sides  of  the  triangle  DKC, 
CE  :  ED  ; :  (2.  6.)  KH  :  HD ;  But  KH=BG, 
and  HD  =  GF  ;  therefore  CE  :  ED  : :  BG  : 
GF  ;  Again,  because  FD  is  parallel  to  EG, 
one  of  the  sides  of  the  triangle  AGE,  ED  ;  DA 
:  :  GF  :  FA  ;  But  it  has  been  proved  that  CE 
:  ED  : :  BG  :  GF;  therefore  the  given  straight  lineAB  is  divided  similarly 
to  AC. 


PROP.  XL    PROB. 

To  find  a  third  proportional  to  two  given  straight  lines. 

Let  AB,  AC  be  the  two  given  straight  lines,  and  let  them  be  placed  so 
as  to  contain  any  angle  ;  it  is  required  to 
find  a  third  proportional  to  AB,  AC. 

Produce  AB,  AC  to  the  points  D,  E  ;  and 
make  BD  equal  to  AC  ;  and  having  joined 
BC,  through  D  draw  DE  parallel  to  it  (Prop. 
31.1.) 

Because  BC  is  parallel  to  DE,  a  side  of 
the  triangle  ADE,  AB  :  (2.  6.)  BD  . :  AC  : 
CE  ;  but  BD=AC:  therefore  AB  :  AC  :; 
AC  :  CE.  Wherefore  to  the  two  given 
straight  lines  AB,  AC  a  third  proportional, 
CE  is  found. 


OF  GEOMETRY.    BOOK  VI. 


131 


PROP.  XII.    PROB. 

To  find  a  fourth  proportional  to  three  given  straight  lines. 

Let  A,  B,  C  be  the  three  given  straight  lines  ;  it  is  required  to  find  a 
fourth  proportional  to  A,  B,  C. 

Take  two  straight  lines  DE,  DF,  containing  any  angle  EDF ;  and  upon 
these  make  DG  equal  to  A,  GE  equal  to  B,  and  DH  equal  to  C  ;  and  hav- 
ing joined  GH,  draw  EF  parallel  (Prop.  31. 1.)  to  it  through  the  point  E. 


And  because  GH  is  parallel  to  EF,  one  of  the  sides  of  the  triangle  DEF, 
DG  :  GE  : :  DH  :  HF  (2.  6.) ;  but  DG=A,  GE=B,  and  DH=C .;  and 
therefore  A  :  B  ; :  0  :  HF.  Wherefore  to  the  three  given  straight  lines, 
A.  B,  C,  a  fourth  proportional  HF  is  foimd. 

PROP.  XIII.     PROB. 
To  find  a  mean  proportional  between  two  given  straight  lines. 

Let  AB,  BC  be  the  two  given  straight  lines  ;  it  is  required  to  find  a  mean 
proportional  between  them. 

Place  AB,  BC  in  a  straight  line,  and  upon  AC  describe  the  semicircle 
ADC,  and  from  the  point  B  (Prop.  11. 
1.)  draw  BD  at  right  angles  to  AC,  and 
join  AD,  DC. 

Because  the  angle  ADC  in  a  semi- 
circle is  a  right  angle  (31.  3.)  and  be- 
cause in  the  right  angled  triangle  ADC, 
DB  is  drawn  from  the  right  angle,  per- 
pendicular to  the  base,  DB  is  a  mean 
proportional  between  AB,  BC,  the  seg- 
ments of  the  base  (Cor.  8.  6.) ;  therefore  between  the  two  given  straight 
lines  AB,  BC,  a  mean  proportional  DB  is  found. 


132  ELEMENTS 


PROP.  XIV.  PROB. 


Equal  parallelograms  which  have  one  angle  of  the  one  equal  to  one  angle  of 
the  other,  have  their  sides  about  the  equal  angles  reciprocally  proportional : 
And  parallelograms  which  have  one  angle  ojf  the  one  equal  to  one  angle  of 
the  other,  and  their  sides  about  the  equal  angles  reciprocally  proportional, 
are  equal  to  one  another. 

Let   AB,   BC    be    equal   parallel- 
ograms, which  have  the  angles  at  B 
equal,  and  let  the  sides  DB,  BE  be 
placed   in  the   same    straight    line ; 
wherefore  also   FB,  BG  are  in  one 
straight  line  (14. 1.) ;  the  sides  of  the 
parallelograms    AB,   BC,   about    the 
equal  angles,  are  reciprocally  propor- 
tional ;  that  is,  DB  is  to  BE,  as  GB  ,^  ^ 
toBF.                                                                                          Ct        C 
Complete  the  parallelogram  FE  ;  and  because  the  parallelograms  AB, 
BC  are  equal,  and  FE  is  another  parallelogram, 

AB:  FE  ::  BC:  FE  (7.5.): 
but  because  the  parallelograms  AB,  FE  have  the  same  altitude, 
AB  :  FE  : :  DB  :  BE  (1.  6.),  also, 
BC  :  FE  : :  GB  :  BF  (1.  6.) ;  therefore 
DB  :  BE  : :  GB  :  BF  (11.  5.).     Wherefore,  the  sides 
of  the  parallelograms  AB,  BC  about  their  equal  angles  are  reciprocally  pro- 
portional. 

But,  let  the  sides  about  the  equal  angles  be  reciprocally  proportional,  viz. 
as  DB  to  BE,  so  GB  to  BF  ;  the  parallelogram  AB  is  equal  to  the  parallel- 
ogram BC. 

Because  DB  :  BE  : :  GB  :  BF,  and  DB  :  BE  :  :  AB  :  FE,  and  GB  : 
BF  :  :  BC  :  EF,  therefore,  AB  :  FE  : :  BC  :  FE  (11.  5.):  wherefore  th© 
parallelogram  AB  is  equal  (9.  5.)  to  the  parallelogram  BC. 

PROP.  XV.     THEOR. 

Equal  triangles  which  have  one  angle  of  the  one  equal  to  one  angle  of  the 
other  have  their  sides  about  the  equal  angles  reciprocally  proportional ;  And 
triangles  which  have  one  angle  in  the  one  equal  to  one  angle  in  the  other, 
and  their  sides  about  the  equal  angles  reciprocally  proportional,  are  equal 
to  one  another. 

Let  ABC,  ADE  be  equal  triangles,  which  have  the  angle  BAC  equal  to 
the  angle  DAE  :  the  sides  about  the  equal  angles  of  the  triangles  are  re- 
ciprocally proportional ;  that  is,  CA  is  to  AD,  as  EA  to  AB. 

Let  the  triangles  be  placed  so  that  their  sides  CA,  AD  be  in  one  straight 
line  ;  wherefore  also  EA  and  AB  are  in  one  straight  line  (14.  1.) ;  join  BD. 
Because  the  triangle  ABC  is  equal  to  the  triangle  ADE,  and  ABD  is  an- 
other triangle ;  therefore,  triangle  CAB  :  triangle  BAD  :  :  triangle  EAD 


OF  GEOMETRY.    BOOK  VI. 


133 


:  triangle  BAD  ;  but  CAB  : 
BAD  :  :  CA  :  AD,  and  EAD  : 
BAD  : ;  EA  :  AB  ;  therefore 
CA:  AD::  EA  :  AB(11.5), 
wherefore  the  sides  of  the  trian- 
gles ABC,  ADE  about  the  equal 
angles  are  reciprocally  propor- 
tional. 

But  let  the  sides  of  the  trian- 
gles ABC,  ADE,  about  the 
equal  angles  be  reciprocally- 
proportional,  viz.  CA  to  AD,  as 
EA  to  AB ;  the  triangle  ABC  is 
equal  to  the  triangle  ADE. 

Having  joined  BD  as  before ;  because  CA  :  AD  :  :  EA  :  AB  ;  and  since 
CA  :  AD  :  :  triangle  ABC  :  triangle  BAD  (1.6.);  and  also  EA  :  AB  :  : 
triangle  EAD  :  triangle  BAD  (11.  5.) ;  therefore,  triangle  ABC  :  triangle 
BAD  : :  triangle  EAD  :  triangle  BAD  ;  that  is,  the  triangles  ABC,  EaD 
have  the  same  ratio  to  the  triangle  BAD  ;  wherefore  the  triangle  ABC  is 
equal  (9.  5.)  to  the  triangle  EAD. 

PROP.  XVI.    THEOR. 

If  four  straight  lines  he  proportionals ,  the  rectangle  contained  hy  the  extremes  is 
equal  to  the  rectangle  contained  hy  the  means;  And  if  the  rectangle  contained 
by  the  extremes  be  equal  to  the  rectangle  contained  hy  the  means ^  the  four 
straight  lines  are  proportionals. 

Let  the  four  straight  lines,  AB,  CD,  E,  F,  be  proportionals,  viz.  as  AB 
to  CD,  so  E  to  F  ;  the  rectangle  contained  by  AB,  F  is  equal  to  the  rect- 
angle contained  by  CD,  E. 

From  the  points  A,  C  draw  (6. 1.)  AG,  CH  at  right  angles  to  AB,  CD  ; 
and  make  AG  equal  to  F,  and  CH  equal  to  E,  and  complete  the  parallel- 
ograms BG,  DH.  Because  AB  :  CD  : :  E  :  F  ;  and  since  E=CH,  and 
F=AG,  AB  :  CD  (7.  5.) :  :  CH  :  AG  ;  therefore  the  sides  of  the  parallel- 
ograms BG,  DH  about  the  equal  angles  are  reciprocally  proportional ;  but 
parallelograms  which  have  their  sides  about  equal  angles  reciprocally  pro- 
portional, are  equal  to  one  another  (14.  6.);  therefore   the  parallelogram 

BG  is  equal  to  the  parallelogram  DH  :       "B] . 

and  the  parallelogram  DGis  contain- 
ed by  the  straight  lines  AB,  F  ;  be-  p- 
cause  AG  is  equal  to  F  ;  and  the  pa- 
rallelogram DH  is  contained  by  CD 
and  E,  because  CH  is  equal  to  E  : 
therefore  the  rectangle  contained  by 
the  straight  lines  AB,  F  is  equal  to  that 
which  is  contained  by  CD  and  E. 

And  if  the  rectangle   contained  by 
the  straight  lines  AB,  F  be  equal  to  that  which  is  contained  by  CD,  E  ; 
these  four  lines  are  proportionals,  viz.  AB  is  to  CD,  as  E  to  F. 


c 


134  ELEMENTS 

The  same  construction  being  made,  because  the  rectangle  contained  by 
the  straight  lines  AB,  F  is  equal  to  that  which  is  contained  by  CD,  E,  and 
the  rectangle  BG  is  contained  by  AB,  F,  because  AG  is  equal  to  F  ;  and 
the  rectangle  DH,  by  CD,  E,  because  CH  is  equal  to  E  ;  therefore  the  pa- 
rallelogram BG  is  equal  to  the  parallelogram  DH,  and  they  are  equiangu- 
lar :  but  the  sides  about  the  equal  angles  of  equal  parallelograms  are  reci- 
procally proportional  (14.  6.) :  wherefore  AB  :  CD  : :  CH  :  AG ;  but  CH 
=E,  and  AG=F;  therefore  AB  :  CD  :  :  E  :  F. 

PROP.  XVn.    THEOR. 

If  three  straight  lines  he  proportionals,  the  rectangle  contained  by  the  extremes  is 
equal  to  the  square  of  the  mean  :  And  if  the  rectangle  contained  by  the  ex- 
tremes he  equal  to  the  square  of  the  mean,  the  three  straight  lines  arepropor- 
tionals. 

Let  the  three  straight  lines,  A,  B,  C  be  proportionals,  viz.  as  A  to  B,  so 
B  to  C  ;  the  rectangle  contained  by  A,  C  is  equal  to  the  square  of  B. 

Take  D  equal  to  B  :  and  because  as  A  to  B,  so  B  to  C,  and  that  B  is 
equal  to  D ;  A  is  (7.  5.)  to  B,  as  D  to  C  :  but  if  four  straight  lines  be  pro- 
portionals, the  rectangle  contained  by  the  extremes  is  equal  to  that  which 
is  contained  by  the  means  (16.  6.) ;  therefore  the 
rectangle  A.C  =  the  rectangle  B.D  ;  but  the  rect-     ^ 
angle  B.D  is  equal  to  the  square  of  B,  because  B=     r?  "~ 
D  ;  therefore   the   rectangle  A.C  is  equal  to  the     i;  ~ 

square  of  B.  ^ 

And  if  the  rectangle  contained  by  A,  C  be  equal  to  the  square  of  B  ;  A  : 
B  : :  B  :  C. 

The  same  construction  being  made,  because  the  rectangle  contained  by 
A,  C  is  equal  to  the  square  of  B,  and  the  square  of  B  is  equal  to  the  rect- 
angle contained  by  B,  D,  because  B  is  equal  to  D  ;  therefore  the  rectangle 
contained  by  A,  C  is  equal  to  that  contained  by  B,  D  ;  but  if  the  rectangle 
contained  by  the  extremes  be  equal  to  that  contained  by  the  means,  the 
four  straight  lines  are  proportionals  (16.  6.) :  therefore  A  :  B  :  :  D  :  C,but 
B=D  ;  wherefore  A  :  B  : :  B  :  C. 

PROP.  XVHL    PROB. 

Upon  a  given  straight  line  to  describe  a  rectilineal  figure  similar,  and  similarly 
situated  to  a  given  rectilineal  figure. 

Let  AB  be  the  given  straight  line,  and  CDEF  the  given  rectilineal  figure 
of  four  sides  ;  it  is  required  upon  the  given  straight  line  AB  to  describe  a 
rectilineal  figure  similar,  and  similarly  situated  to  CDEF. 

Join  DF,  and  at  the  points  A,  B  in  the  straight  line  AB,  make  (Prop.  23. 
1.)  the  angle  BAG  equal  to  the  angle  atC,  and  the  angle  ABG  equal  to 
the  angle  CDF  ;  therefore  the  remaining  angle  CFD  is  equal  to  the  re- 
maining angle  AGB  (4.  Cor.  32. 1.) :  wherefore  the  triangle  FCD  is  equi- 
angular to  the  triangle  GAB  :  Again,  at  the  points  G,  B  in  the  straight 
line  GB  make  (Prop.  23. 1.)  the  angle  BGH  equal  to  the  angle  DFE,  and 
the  angle  GBH  equal  to  FDE  ;  therefore  the  remaining  angle  FED  is 


OF  GEOMETRY.    BOOK  VI. 


135 


equal  to  the  remaining  angle  GHB,  and  the  triangle  FDE  equiangular  to 
the  triangle  GBH  :  then,  because  the  angle  AGB  is  equal  to  the  angle 
CFD,  BGH  to  DFE  the  whole  angle  AGH  is  equal  to  the  whole  CF£  : 


L      F 


for  the  same  reason,  the  angle  ABH  is  equal  to  the  angle  ODE  ;  also  the 
angle  at  A  is  equal  to  the  angle  at  C,  and  the  angle  GHB  to  FED  ;  There- 
fore the  rectilineal  figure  ABHG  is  equiangular  to  CDEF :  but  likewise 
these  figurfcs  have  their  sides  about  the  equal  angles  proportionals  :  for  the 
triangles  GAB,  FCD  being  equiangular, 

BA  :  AG  : :  DC  :  CF  (4.  6.) ;  for  the  same  reason, 
AG  :  GB  : :  CF  :  FD ;  and  because  of  the  equian- 
gular triangles  BGH,  DFE,  GB  :  GH  : :  FD  :  FE  ;  therefore, 

ex  eequali  (22.  5.)  AG  :  GH  : :  CF  :  FE. 
In  the  same  manner,  it  may  be  proved,  that 

AB  :  BH  : :  CD  :  DE.     Also  (4.  6.), 
GH  :  HB  : :  FE  :  ED.     Wherefore,  because  the  rectili- 
neal figures  ABHG,  CDEF  are  equiangular,  and  have  their  sides  about 
the  equal  angles  proportionals,  they  are  similar  to  one  another  (def.  1.  6.). 

Next,  Let  it  be  required  to  describe  upon  a  given  straight  line  AB,  a 
rectilineal  figure  similar,  and  similarly  situated  to  the  rectilineal  figure 
CDKEF. 

Join  DE,  and  upon  the  given  straight  line  AB  describe  the  rectilineal 
figure  ABHG  similar,  and  similarly  situated  to  the  quadrilateral  figure 
CDEF,  by  the  former  case  ;  and  at  the  points  B,  H  in  the  straight  line 
BH,  make  the  angle  HBL  equal  to  the  angle  EDK,  and  the  angle  BHL 
equal  to  the  angle  DEK  ;  therefore  the  remaining  angle  at  K  is  equal  to 
the  remaining  angle  at  L ;  and  because  the  figures  ABHG,  CDEF  are 
similar,  the  angle  GHB  is  equal  to  the  angle  FED,  and  BHL  is  equal  to 
DEK  ;  wherefore  the  whole  angle  GHL  is  equal  to  the  whole  angle  FEK ; 
for  the  same  reason  the  angle  ABL  is  equal  to  the  angle  CDK  :  therefore 
the  five-sided  figures  AGHLB,  CFEKD  are  equiangular ;  and  because 
the  figures  AGHB,  CFED  are  similar,  GH  is  to  HB  as  FE  to  ED  ;  and 
as  HB  to  HL,  so  is  ED  to  EK  (4.  6.) ;  therefore,  ex  eequali  (22.  5.),  GH 
is  to  HL,  as  FE  to  EK :  for  the  same  reason,  AB  is  to  BL,  as  CD  to  DK : 
and  BL  is  to  LH,  as  (4.  6.)  DK  to  KE,  because  the  triangles  BLH,  DKE 
are  equiangular  :  therefore,  because  the  five-sided  figures  AGHLB. 
CFEKD  are  equiangular,  and  have  their  sides  about  the  equal  angles  pro- 
portionals, they  are  similar  to  one  another  ;  and  in  the  same  manner  a  rec- 


136 


ELEMENTS 


tilineal  figure  of  six,  or  more,  sides  may  be  described  upon  a  given  straight 
line  similar  to  one  given,  and  so  on. 

PROP.  XIX.    THEOR. 


Similar  triangles  are  to  one  another  tn  the  duplicate  ratio  of  the  homologous 

sides. 

Let  ABC,  DEF  be  simi- 
lar triangles,  having  the  an- 
gle B  equal  to  the  angle  E, 
and  let  AB  be  to  BC,  as 
DE  to  EF,  so  that  the  side 
BC  is  homologous  to  EF 
(def.  13.  5.)  :  the  triangle 
ABC  has  to  the  triangle 
DEF,  the  duplicate  ratio 
of  that  which  BC  has  to 
EF. 

Take  BG  a  third  proportional  to  BC  and  EF  (11.  6.),  or  such  that 
BC  :  EF  : :  EF  :  BG,  and  join  GA.     Then,  because 
AB  :  BC  : :  DE  :  EF,  alternately  (16.  5.), 
AB  :  DE  : :  BC  :  EF ;  but 
BC  :  EF  : :  EF  :  BG ;  therefore  (11.  5.) 
AB  :  DE  ::  EF  :  BG;  wherefore  the  sides  of  the  triangles 
ABG,  DEF,  which  are  about  the  equal  angles,  are  reciprocally  propor- 
tional ;  but  triangles,  which  have  the  sides  about  two  equal  angles  recipro- 
cally proportional,  are  equal  to 

one  another  (15. 6.) :  therefore  A 

the  triangle  ABG  is  equal  to 

thetriangle  DEF;  and  because  y/       \  -w^ 

that  BC  is  to  EF,  as  EF  to  X  /       \      -  1> 

BG ;  and  that  if  three  straight 
lines  be  proportionals,  the  first 
has  to  the  third  the  duplicate 

ratio  of  that  which  it  has  to  the  

second ;  BC  therefore  has  to      B         G  C  33  F 

BG  the  duplicate  ratio  of  that  which  BC  has  to  EF.  But  as  BC  to  BG, 
so  is  (1.  6.)  the  triangle  ABC  to  the  triangle  ABG  :  therefore  the  triangle 
ABC  has  to  the  triangle  ABG  the  duplicate  ratio  of  that  which  BC  has  to 
EF  :  and  the  triangle  ABG  is  equal  to  the  triangle  DEF  ;  wherefore  also 
the  triangle  ABC  has  to  the  triangle  DEF  the  duplicate  ratio  of  that  which 
BC  has  to  EF. 

Cor.  From  this,  it  is  manifest,  that  if  three  straight  lines  be  propor- 
tionals, as  the  first  is  to  the  third,  so  is  any  triangle  upon  the  first  to  a 
similar,  and  similarly  described  triangle  upon  the  second. 


OF  GEOMETRY.     BOOK  VI.  137 


PROP.  XX.    THEOR. 

Similar  polygons  may  he  divided  into  the  same  number  of  similar  triangles,  hav- 
ing the  same  ratio  to  one  another  that  the  polygons  have ;  and  the  polygons 
have  to  one  another  the  duplicate  ratio  of  that  which  their  homologous  sides 
have. 

Let  ABODE,  FGHKL,  be  similar  polygons,  and  let  AB  be  the  homo- 
logous side  to  FG:  the  polygons  ABODE,  FGHKL,  may  be  divided  into 
the  same  number  of  similar  triangles,  whereof  each  has  to  each  the  same 
ratio  which  the  polygons  have  ;  and  the  polygon  ABODE  has  to  the  poly- 
gon FGHKL  a  ratio  duplicate  of  that  which  the  side  AB  has  to  the  side 
FG. 

Join  BE,  EO,  GL,  LH  :  and  because  the  polygon  ABODE  is  similar 
to  the  polygon  FGHKL,  the  angle  BAE  is  equal  to  the  angle  GFL  (def. 
L  6.),  and  BA  :  AE  : :  GF  :  FL  (def.  1.6.):  wherefore,  because  the  tri- 
angles ABE,  FGL  have  an  angle  in  one  equal  to  an  angle  in  the  other, 
and  their  sides  about  these  equal  angles  proportionals,  the  triangle  ABE  is 
equiangular  (6.  6.),  and  therefore  similar,  to  the  triangle  FGL  (4.  6.) : 
wherefore  the  angle  ABE  is  equal  to  the  angle  FGL  :  and,  because  the 
polygons  are  similar,  the  whole  angle  ABO  is  equal  (def.  1.  6.)  to  the  whole 
angle  FGH  ;  therefore  the  remaining  angle  EBO  is  equal  to  the  remain- 
ing angle  LGH  :  now  because  the  triangles  ABE,  FGL  are  similar, 

EB  :  BA  :  :  LG  :  GF ;  and  also  because  the 
polygons  are  similar,  AB  :  BO  :  :  FG  ;  GH  (def.  1.6.);  therefore,  ex 
aequali  (22.  5.)  EB  :  BO  :  :  LG  ;  GH,  that  is,  the  sides  about  the  equal 
angles  EBO,  LGH  are  proportionals ;  therefore  (6.  6.)  the  triangle  EBO 


is  equiangular  to  the  triangle  LGH,  and  similar  to  it  (4.  6.).  For  the 
same  reason,  the  triangle  EOD  is  likewise  similar  to  the  triangle  LHK  ; 
therefore  the  similar  polygons  ABODE,  FGHKL  are  divided  into  the  same 
number  of  similar  triangles. 

Also  these  triangles  have,  each  to  each,  the  same  ratio  which  the  poly- 
gons have  to  one  another,  the  antecedents  being  ABE,  EBO,  EOD,  and 
the  consequents  FGL,  LGH,  LHK  :  and  the  polygon  ABODE  has  to  the 
polygon  FGHKL  the  duplicate  ratio  of  that  which  the  side  AE  has  to  the 
homologous  side  FG. 

Because  the  triangle  ABE  is  similar  to  the  triangle  FGL,  ABE  has  to 
FGL  the  duplicate  ratio  (19.  6.)  of  that  which  the  side  BE  has  to  the  side 

18 


138  ELEMENTS 

GL :  for  the  same  reason,  the  triangle  BEC  has  to  GLH  the  duplicate 
ratio  of  that  which  BE  has  to  GL  :  therefore,  as  the  triangle  ABE  to  the 
triangle  FGL,  so  (11.  5.)  is  the  triangle  BEC  to  the  triangle  GLH.  Again, 
because  the  triangle  EBC  is  similar  to  the  triangle  LGH,  EBC  has  to 
LGH  the  duplicate  ratio  of  that  which  the  side  EC  has  to  the  side  LH  : 
for  the  same  reason,  the  triangle  ECD  has  to  the  triangle  LHK,  the  du- 
plicate ratio  of  that  which  EC  has  to  LH  :  therefore,  as  the  triangle  EBC 
to  the  triangle  LGH,  so  is  (11.  5.)  the  triangle  ECD  to  the  triangle  LHK  : 
but  it  has  been  proved,  that  the  triangle  EBC  is  likewise  to  the  triangle 
LGH,  as  the  triangle  ABE  to  the  triangle  FGL.  Therefore,  as  the  trian- 
gle ABE  is  to  the  triangle  FGL,  so  is  the  triangle  EBC  to  the  triangle 
LGH,  and  the  triangle  ECD  to  the  triangle  liHK  :  and  therefore,  as  one 
of  the  antecedents  to  one  of  the  consequents,  so  are  all  the  antecedents  to 
all  the  consequents  (12.  5.).     Wherefore,  as  the  triangle  ABE  to  the  tri- 


angle FGL,  so  is  the  polygon  ABCDE  to  the  polygon  FGHKL :  but  the 
triangle  ABE  has  to  the  triangle  FGIi,  the  duplicate  ratio  of  that  which 
the  side  AB  has  to  the  homologous  side  FG.  Therefore  also  the  polygon 
ABCDE  has  to  the  polygon  FGHKL  the  duplicate  ratio  of  that  which 
AB  has  to  the  homologous  side  FG. 

CoR.  L  In  like  manner  it  maybe  proved,  that  similar  figures  of  four 
sides,  or  of  any  number  of  sides,  are  one  to  another  in  the  duplicate  ratio  of 
their  homologous  sides,  and  the  same  has  already  been  proved  of  triangles  : 
therefore,  universally,  similar  rectilineal  figures  are  to  one  another  in  the 
duplicate  ratio  of  their  homologous  sides. 

CoR.  2.  And  if  to  AB,  FG,  two  of  the  homologous  sides,  a  third  pro- 
portional M  be  taken,  AB  has  (def.  IL  5.)  to  M  the  duplicate  ratio  of  that 
which  AB  has  to  FG  :  but  the  four-sided  figure,  or  polygon,  upon  AB  has 
to  th3  four-sided  figure,  or  polygon,  upon  FG  likewise  the  duplicate  ratio 
of  that  which  AB  has  to  FG :  therefore,  as  AB  is  to  M,  so  is  the  figure 
upon  AB  to  the  figure  upon  FG,  which  was  also  proved  in  triangles  (Cor. 
19.  6.).  Therefore,  universally,  it  is  manifest,  that  if  three  straight  lines 
be  proportionals,  as  the  first  to  the  third,  so  is  any  rectilineal  figure  upon 
the  first,  to  a  similar,  and  similarly  described  rectilineal  figure  upon  the  se- 
cond. 

Cor.  3.  Because  all  squares  are  similar  figures,  the  ratio  of  any  two 
squares  to  one  another  is  the  same  with  the  duplicate  ratio  of  their  sides ; 
and  hence,  also,  any  two  similar  rectilineal  figures  are  to  one  another  as  the 
squares  of  their  homologous  sides. 


OF  GEOMETRY.    BOOK  VI.  139 


SCHOLIUM. 


If  two  polygons  are  composed  of  the  same  number  of  triangles  similar, 
and  similarly  situated,  those  two  polygons  will  be  similar. 

For  the  similarity  of  the  two  triangles  will  give  the  angles  EAB=LFG, 
ABE=FGL,EBC=LGH;  hence,  ABC =FGH,  likewise  BCD=GHK, 
&c.  Moreover,  we  shall  have,  EA  :  LF  :  :  AB  :  FG  :  :  EB  :  LG  :  :  BC 
:  GH,  &-C. ;  hence  the  two  polygons  have  their  angles  equal  and  their  sides 
proportional ;  consequently  they  are  similar. 

PROP.  XXI     THEOR. 

Rectilineal  figures  which  are  similar  to  the  same  rectilineal  figure  j  are  also 
similar  to  one  another. 

Let  each  of  the  rectilineal  figures  A,  B  be  similar  to  the  rectilineal  figure 
C  :  The  figure  A  is  similar  to  the  figure  B. 

Because  A  is  similar  to  C,  they  are  equiangular,  and  also  have  their 
sides  about  the  equal  angles  proportionals  (def.  1.  6.).  Again,  because  B 
is  similar  to  C,  they  are  equiang-ular,  and  have  their  sides  about  the  equal 
angles  proportionals  (def.  1.  6.) :  therefore  the  figures  A,  B,  are   each  of 


them  equiangular  to  C,  and  have  the  sides  about  the  equal  angles  of  each 
of  them,  and  of  C,  proportionals.  Wherefore  the  rectilineal  figures  A  and 
B  are  equiangular  (1.  Ax.  1.),  and  have  their  sides  about  the  equal  angles 
proportionals  (11.  5.).     Therefore  A  is  similar  (def.  1.  6.)  to  B. 

PROP.  XXII     THEOR. 

If  four  straight  lines  he  proportionals,  the  similar  rectilineal  figures  similarly 
described  upon  them  shall  also  he  proportionals ;  and  if  the  similar  rectilineal 
figures  similarly  described  upon  four  straight  lines  he  proportionals ,  those 
straight  lines  shall  be  proportionals. 

Let  the  four  straight  lines,  AB,  CD,EF,  GH  be  proportionals,  viz.  AB 
to  CD,  as  EF  to  GH,  and  upon  AB,  CD  let  the  similar  rectilineal  figures 
KAB,  LCD  be  similarly  described  ;  and  upon  EF,  GH  the  similar  recti- 
lineal figures  MF,  NH,  in  like  manner :  the  rectilineal  figure  KAB  is  to 
LCD,  as  MF  to  NH. 

To  AB,  CD  take  a  third  proportional  (11.  6.)  X ;  and  to  EF,  GH,a 
third  proportional  0  ;  and  because 


140 


ELEMENTS 


AB  :  CD .:  :  EF  :  GH,  and 

CD  :  X  :  :  GH  :  (11.  5.)  O,  ex  aequali  f22.  5.) 

AB  :  X  :  :  EF  :  0.     But 

AB  :  X  (2.  Cor.  20.  6.)  :  :  KAB  :  LCD  ;  and 

EF  :  0  :  :  (2.  Cor.   20.   6.)  MF  :  NH  ;   therefore 
KAB  :  LCD  (2.  Cor.  20.  6.)  :  :  MF  :  NH. 

And  if  the  figure  KAB  be  to  the  figure  LCD,  as  the  figure  MF  to  the 
figure  NH,  AB  is  to  CD,  as  EFto  GH. 

Make  (12.  6.)  as  AB  to  CD,  so  EF  to  PR,  and  upon  PR  describe  (18. 
6.)  the  rectilineal  figure  SR  similar,  and  similarly  situated  to  either  of  the 


E  T'       G        H       O        P         R 

figures  MF,  NH  :  then,  because  that  as  AB  to  CD,  so  is  EF  to  PR,  and 
upon  AB,  CD  are  described  the  similar  and  similarly  situated  rectilineals 
KAB,  LCD,  and  upon  EF,  PR,  in  like  manner,  the  similar  rectilineals 
MF,  SR  ;  KAB  is  to  LCD,  as  MF  to  SR  ;  but  by  the  hypothesis,  KAB 
is  to  LCD,  as  MF  to  NH ;  and  therefore  the  rectilineal  MF  having  the 
same  ratio  to  each  of  the  two  NH,  SR,  these  two  are  equal  (9.  5.)  to  one 
another ;  they  are  also  similar,  and  similarly  situated  ;  therefore  GH  is 
equal  to  PR  :  and  because  as  AB  to  CD,  so  is  EF  to  PR,  and  because  PR 
is  equal  to  GH,  AB  is  to  CD,  as  EF  to  GH. 


PROP.  XXHL    THEOR. 

Equiangular  parallelograms  have  to  one  another  the  ratio  which  is  compounded 
of  the  ratios  of  their  sides. 

Let  AC,  CF  be  equiangular  parallelograms  having  the  angle  BCD 
equal  to  the  angle  ECG ;  the  ratio  of  the  parallelogram  AC  to  the  paral- 
lelogram CF,  is  the  same  with  the  ratio  which  is  compounded  of  the  ratios 
of  their  sides. 

Let  BC,  CG  be  placed  in  a  straight  line  ;  therefore  DC  and  CE  are  also 
in  a  straight  line  (14. 1.);  complete  the  parallelogram  DG  ;  and,  taking 
any  straight  line  K,  make  (12.  6.)  as  BC  to  CG,  so  K  to  L ;  and  as  DC 
to  CE,  so  make  (12.  6.)  L  to  M  :  therefore  the  ratios  of  K  to  L,  and  L  to 
M,  are  the  same  with  the  ratios  of  the  sides,  viz.  of  BC  to  CG,  and  of  DC 
to  CE.  But  the  ratio  of  K  to  M,  is  that  which  is  said  to  be  compounded 
(def.  10. 5.)  of  the  ratios  of  K  to  L,  and  L  to  M  ;  wherefore  also  K  has  to 


OF  GEOMETRY.    BOOK  VI. 


141 


K  I 


M  the  ratio  compounded  of  the  ratios  of 
the  sides  of  the  parallelograms.  Now, 
because  as  BC  to  CG,  so  is  the  parallel- 
ogram AC  to  the  parallelogram  CH  (1. 
6.) ;  and  as  BC  to  CG,  so  is  K  to  L ; 
therefore  K  is  (11.  5.)  to  L,  as  the  paral- 
lelogram AC  to  the  parallelogram  CH : 
again,  because  as  DC  to  CE,  so  is  the 
parallelogram  CH  to  the  parallelogram 
CF :  and  as  DC  to  CE,  so  is  L  to  M ; 
therefore  L  is  (1 1.  5.)  to  M,  as  the  paral- 
lelogram CH  to  the  parallelogram  CF  : 
therefore,  since  it  has  been  proved,  that 
as  K  to  L,  so  is  the  parallelogram  AC 
to  the  parallelogram  CH  ;  and  as  L  to  M,  so  the  parallelogram  CH  to  the 
parallelogram  CF  ;  ex  ajquali  (22.  5.),  K  is  to  M,  as  the  parallelogram 
AC  to  the  parallelogram  CF ;  but  K  has  to  M  the  ratio  which  is  com- 
pounded of  the  ratios  of  the  sides ;  therefore  also  the  parallelogram  AC 
has  to  the  parallelogram  CF  the  ratio  which  is  compounded  of  the  ratios 
of  the  sides. 

CoR.  Hence,  any  two  rectangles  are  to  each  other  as  the  products  of 
their  bases  multiplied  hy  their  altitudes. 

SCHOLIUM. 

Hence  the  product  of  the  base  by  the  altitude  may  be  assumed  as  the 
measure  of  a  rectangle,  provided  we  understand  by  this  product  the  pro- 
duct of  two  numbers,  one  of  which  is  the  number  of  linear  units  contained 
in  the  base,  the  other  the  number  of  linear  units  contained  in  the  altitude. 

Still  this  measure  is  not  absolute  but  relative  :  it  supposes  that  the  area 
of  any  other  rectangle  is  computed  in  a  similar  manner,  by  measuring  its 
sides  with  the  same  linear  unit ;  a  second  product  is  thus  obtained,  and 
the  ratio  of  the  two  products  is  the  same  as  that  of  the  two  rectangles, 
agreeably  to  the  proposition  just  demonstrated. 

For  example,  if  the  base  of  the  rectangle  A  contained  three  units,  and  its 
altitude  ten,  that  rectangle  will  be  represented  by  the  number  3x10,  or 
30,  a  number  which  signifies  nothing  while  thus  isolated  ;  but  if  there  is  a 
second  rectangle  B,  the  base  of  which  contains  twelve  units,  and  the  alti- 
tude seven,  this  rectangle  would  be  represented  by  the  number  12  X  7=84  ; 
and  we  shall  hence  be  entitled  to  conclude  that  the  two  rectangles  are  to 
each  other  as  30  is  to  84  ;  and  therefore,  if  the  rectangle  A  were  to  be  as- 
sumed as  the  unit  of  measurement  in  surfaces,  the  rectangle  B  would  then 
have  1^  for  its  absolute  measure  ;  or,  which  amounts  to  the  same  thing,  it 
would  be  equal  to  |^  of  a  superficial  unit. 

It  is  more  common  and  more  simple  to  assume  the  squares  as  the  unit  of 
surface  ;  and  to  select  that  square  whose  side  is  the  unit  of  length.  In 
this  case,  the  measurement  which  we  have  regarded  merely  as  relative^ 
becomes  absolute  :  the  number  30,  for  instance,  by  which  the  rectangle  A 
was  measured,  now  represents  30  superficial  units,  or  30  of  those  squares, 
which  have  each  of  their  sides  equal  to  unity. 


142  ELEMENTS 

Cor.  1.  Hence,  the  area  of  anyparallclogram  is  equal  to  the  product  of 
its  base  by  its  altitude. 

Cor.  2.  It  likewise  follows,  that  the  area  of  any  triangle  is  equal  to  the 
product  of  its  base  by  half  its  altitude. 

PROP.  XXIV.    THEOR. 

The  parallelograms  about  the  diameter  of  any  parallelogram,  are  similar  to  the 
whole,  and  to  one  another. 

Let  ABCD  be  a  parallelogram,  of  which  the  diameter  is  AC  ;  and  EG, 
HK  the  parallelograms  about  the  diameter:  the  parallelograms  EG,  HK 
are  similar,  both  to  the  whole  parallelogram  ABCD,  and  to  one  another. 

Because  DC,  GF  are  parallels,  the  angle  ADC  is  equal  (29.  L)  to  the 
angle  AGF  :  for  the  same  reason,  because  BC,  EF  are  parallels,  the  an- 
gle ABC  is  equal  to  the  angle  AEF  :  and  each  of  the  angles  BCD,  EFG 
is  equal  to  the  opposite  angle  DAB  (34.  1*),  and  therefore  are  equal  to  one 
another,  wherefore  the  parallelograms  ABCD,  AEFG  are  equiangular 
And  because  the  angle  ABC  is  equal  to  the  angle  AEF,  and  the  angle 
BAC  common  to  the  two  triangles  BAG, 
EAF,  they  are  equiangular  to  one  another  ; 
therefore  (4.  6.)  as  AB  to  BC,  so  is  AE  to 
EF  ;  and  because  the  opposite  sides  of  paral- 
lelograms are  equal  to  one  another  (34.  1.), 
AB  is  (7.  5.)  to  AD,  as  AE  to  AG ;  and  DC 
to  CB,  as  GF  to  FE ;  and  also  CD  to  DA, 
as  FG  to  GA :  therefore  the  sides  of  the  pa- 
rallelograms ABCD,  AEFG  about  the  equal 
angles  are  proportionals ;  and  they  are 
therefore  similar  to  one  another  (def.  1.6.);  for  the  same  reason,  the  pa- 
rallelogram ABCD  is  similar  to  the  parallelogram  FHCK.  Wherefore 
each  of  the  parallelograms,  GE,  KH  is  similar  to  DB  :  but  rectilineal 
figures  which  are  similar  to  the  same  rectilineal  figure,  are  also  similar  to 
one  another  (21.  6.) ;  therefore  the  parallelogram  GE  is  similar  to  KH. 

PROP.  XXV.     PROB. 

To  describe  a  rectilineal  figure  which  shall  be  similar  to  one,  and  equal  to 
another  given  rectilineal  figure. 

Let  ABC  be  the  given  rectilineal  figure,  to  which  the  figure  to  be  de- 
scribed is  required  to  be  similar,  and  D  that  to  which  it  must  be  equal.  It 
is  required  to  describe  a  rectilineal  figure  similar  to  ABC,  and  equal  to  D. 

Upon  the  straight  line  BC  describe  (Cor.  Prop.  45.  1.)  the  parallelogram 
BE  equal  to  the  figure  ABC  ;  also  upon  CE  describe  (Cor.  Prop.  45.  1.) 
the  parallelogram  CM  equal  to  D,  and  having  the  angle  FCE  equal  to  the 
angle  CBL:  therefore  BC  and  CF  are  in  a  straight  line  (29.  l.or  14.1.),  as 
also  LE  and  EM  ;  between  BC  and  CF  find  (13.  6.)  a  mean  proportional 
GH,  and  upon  GH  describe  (18.  6.)  the  rectilineal  figure  KGH  similar, 
and  similarly  situated,  to  the  figure  ABC.     And  because  BC  is  to  GH  as 


OF  GEOxMETRY.    BOOK  VI. 


143 


GH  to  CF,  and  if  three  straight  lines  be  proportionals,  as  the  first  is  to  the 
third,  so  is  (2.  Cor.  20.  6.)  the  figure  upon  the  first  to  the  similar  and  simi- 
larly described  figure  upon  the  second  ;  therefore  as  BC  to  CF,  so  is  the 


figure  ABC  to  the  figure  KGH  :  but  as  BC  to  CF,  so  is  (1.  6.)  the  paral- 
lelogram BE  to  the  parallelogram  EF  :  therefore  as  the  figure  ABC  is  to 
the  figure  KGH,  so  is  the  parallelogram  BE  to  the  parallelogram  EF  (11. 
5.) :  but  the  rectilineal  figure  ABC  is  equal  to  the  parallelogram  BE  ;  there- 
fore the  rectilineal  figure  KGH  is  equal  (14.  5.)  to  the  parallelogram  EF  : 
but  EF  is  equal  to  the  figure  D  ;  wherefore  also  KGH  is  equal  to  D  ;  and 
it  is  similar  to  ABC.  Therefore  the  rectilineal  figure  KGH  has  been  de- 
scribed similar  to  the  figure  ABC,  and  equal  to  D. 


PROP.  XXVI.    THEOR. 

If  two  similar 'parallelograms  have  a  common  angle,  and  he  similarly  situated^ 
they  are  about  the  same  diameter. 

Let  the  parallelograms  ABCD,  AEFG  be  similar  and  similarly  situated, 
and  have  the  angle  DAB  common ;  ABCD  and  AEFG  are  about  the 
same  diameter. 

For,  if  not,  let,  if  possible,  the  parallelogram 
BD  have  its  diameter  AHC  in  a  different 
straight  line  from  AF,  the  diameter  of  the  pa- 
rallelogram EG,  and  let  GF  meet  AHC  in  H  ; 
and  through  H  draw  HK  parallel  to  AD  or 
BC  ;  therefore  the  parallelograms  ABCD, 
AKHG  being  about  the  same  diameter,  are 
similar  to  one  another  (24.  6.) :  wherefore,  as 
DA  to  AB,  so  is  (def.  1.  6.)  GA  to  AK;  but 
because  ABCD  and  AEFG  are  similar  paral- 
lelograms', as  DA  is  to  AB,  so  is  GA  to  AE  ;  therefore  (11.  5.)  as  Gk  to 
AE,  so  GA  to  AK  ;  wherefore  GA  has  the  same  ratio  to  each  of  the  straight 
lines  AE,  AK  ;  and  consequently  AK  is  equal  (9.  5.)  to  AE,  the  less  to 
the  greater,  which  is  impossible ;  therefore  ABCD  and  AKHG  are  not 
about  the  same  diameter  ;  wherefore  ABCD  and  AEFG  must  be  about 
the  same  diameter. 


144 


ELEMENTS 


PROP.  XXVII.    THEOR. 

Of  all  the  rectangles  contained  hy  the  segments  of  a  given  straight  line,  the 
greatest  is  the  square  which  is  described  on  half  the  line. 

Let  AB  be  a  given  straight  line,  which  is  bisected  in  C  ;  and  let  D  be 

any  point  in  it,  the  square  on  AC  is  greater 

than  the  rectangle  AD,  DB.  ^  C       D       B 

For,  since  the  straight  line  AB  is  divided  into  two  equal  parts  in  C,  and 
into  two  unequal  parts  in  D,  the  rectangle  contained  by  AD  and  DB,  to- 
gether with  the  square  of  CD,  is  equal  to  the  square  of  AC  (5.  2.).  The 
square  of  AC  is  therefore  greater  than  the  rectangle  AD.DB. 

PROP.  XXVIII.     PROB. 

To  divide  a  given  straight  line,  so  that  the  rectangle  contained  by  its  segments 
may  be  equal  to  a  given  space  ;  but  that  space  must  not  be  greater  than  the 
square  of  half  the  given  line. 

Let  AB  be  the  given  straight  line,  and  let  the  square  upon  the  given 
straight  line  C  be  the  space  to  which  the  rectangle  contained  by  the  seg- 
ments of  AB  must  be  equal,  and  this  square,  by  the  determination,  is  not 
greater  than  that  upon  half  the  straight  line  AB. 

Bisect  AB  in  D,  and  if  the  square  upon  AD  be  equal  to  the  square  upon 
C,  the  thing  required  is  done  :  But  if  it  be  not  equal  to  it,  AD  must  be 
greater  than  C,  according  to  the  deter- 
mination :  Draw  DE  at  right  angles  to 
AB,  and  make  it  equal  to  C :  produce 
ED  to  F,  so  that  EF  be  equal  to  AD 
or  DB,  and  from  the  centre  E,  at  the 
distance  EF,  describe  a  circle  meeting 
AB  in  G.  Join  EG;  and  because  AB 
is  divided  equally  in  D,  and  unequally 
in  G,  AG.GB-f  DG2=(5.2.)  DB2= 
EG2.  But  (47.  1.)  ED2+DG2=EG2;  therefore,  AG. GB  +  DG2=ED2 
4-DG2,  and  taking  away  DG^,  AG.GB=ED2.  Now  ED=C,  therefore 
the  rectangle  AG.GB  is  equal  to  the  square  of  C  :  and  the  given  line  AB 
is  divided  in  G,  so  that  the  rectangle  contained  by  the  segments  AG,  GB 
is  equal  to  the  square  upon  the  given  straight  line  C. 


&  B 


PROP.  XXIX.     PROB. 

To  produce  a  given  straight  line,  so  that  the  rectangle  contained  by  the  segments 
between  the  extremities  of  the  given  line,  and  the  points  to  which  it  is  pro- 
duced, may  be  equal  to  a  given  space. 

Let  AB  be  the  given  straight  line,  and  let  the  square  upon  the  given 
straight  line  C  be  the  space  to  which  the  rectangle  under  the  segments  of 
AB  produced,  must  be  equal. 


OF  GEOMETRY.    BOOK  VI. 


145 


Bisect  AB  in  D,  and  draw  BE  at  right  angles  to  it,  so  that  BE  be  equal 
to  C ;  and  having  joined  DE,  from  the  centre  D  at  the  distance  DE  de- 
scribe'a  circle  meeting  AB  produced  in  G. 
And  because  AB  is  bisected  in  D,  and 
produced  to   G,  (6.  2.)  AG.GB+DB2= 
DG2=DE2. 

But  (47.  1.)  DE2=DB2+BE2,  there- 
fore AG.GB  -f  DB2  =  DB2  -f  BE2,  and 
AG.GB=BE2.  Now,  BE  =  C  ;  where- 
fore the  straight  line  AB  is  produced  to 
G,  so  that  the  rectangle  contained  by  the 
segments  AG,  GB  of  the  line  produced, 
is  equal  to  the  square  of  C. 


PROP.  XXX.    PROB. 


To  cut  a  given  straight  line  in  extreme  and  mean  ratio. 

Let  AB  be  the  given  straight  line ;  it  is  required  to  cut  it  in  extreme  and 
mean  ratio. 

Upon  AB  describe  (Prop.  46.  1.)  the  square  BC,  and  produce  CA  taD, 
so  that  the  rectangle  CD. DA  may  be  equal  to  the  square  CB  (29.  6.). 
Take  AE  equal  to  AD,  and  complete  the  rectangle  DF  under  DC  and 
AE,  or  under  DC  and  DA.  Then,  because  the 
rectangle  CD.DA  is  ^qual  to  the  square  CB,  the 
rectangle  DF  is  equal  to  CB.  Take  away  the 
common  part  CE  from  each,  and  the  remainder 
FB  is  equal  to  the  remainder  DE.  But  FB  is 
the  rectangle  contained  by  FE  and  EB,  that  is, 
by  AB  and  BE ;  and  DE  is  the  square  upon  AE ; 
therefore  AE  is  a  mean  proportional  between 
AB  and  BE  (17.  6.),  or  AB  is  to  AE  as  AE  to  EB. 
But  AB  is  greater  than  AE ;  wherefore  AE  is 
greater  than  EB  (14.  5.):  Therefore  the  straight 
line  AB  is  cut  in  extreme  and  mean  ratio  in  E  (def. 
3.  6.). 

Otherwise. 

Let  AB  be  the  given  straight  line ;  it  is  required  to  cut  it  in  extreme 
and  mean  ratio. 

Divide  AB  in  the  point  C,  so  that  the  rectangle  contained  by  AB,  BC 
be  equal  to  the  square  of  AC  (11.  2.):  Then  be- 
cause the  rectangle  AB.BC  is  equal  to  the  square     ^ n g 

of  AC,  as  BA  to  AC,  so  is  AC  to   CB  (17.  6.) ; 
Therefore  AB  is  cut  in  extreme  and  mean  ratio  in  C  fdef.  3.  6.). 

19 


146 


ELEMENTS 


PROP.  XXXI.     THEOR. 

In  right  angled  triangles ^  the  rectilineal  figure  described  upon  the  side  oppcf 
site  to  the  right  angle,  is  equal  to  the  similar^  and  similarly/  described 
figures  upon  the  sides  containing  the  right  angle. 

Let  ABC  be  a  right  angled  triangle,  having  the  right  angle  BAG  :  The 
rectilineal  figure  described  upon  BC  is  equal  to  the  similar,  and  similarly 
described  figures  upon  BA,  AC. 

Draw  the  perpendicular  AD  ;  therefore,  because  in  the  right  angled  tri- 
angle ABC,  AD  is"  drawn  from  the  right  angle  at  A  perpendicular  to  the 
base  BC,  the  triangles  ABD,  ADC  are  similar  to  the  whole  triangle  ABC, 
and  to  one  another  (8.  6.),  and  because  the  triangle  ABC  is  similar  to 
ADB,  as  CB  to  BA,  so  is  BA  to  BD  (4.  6.) ;  and  because  these  three 
straight  lines  are  proportionals,  as  the  first  to  the  third,  so  is  the  figure  upon 
the  first  to  the  similar,  and  similarly  described  figure  upon  the  second  (2. 
Cor.  20.  6.) :  Therefore,  as  CB  to  BD, 
so  is  the  figure  upon  CB  to  the  similar 
and  similarly  described  figure  upon 
BA :  and  inversely  (B.  5.),  as  DB  to 
BC,  so  is  the  figure  upon  BA  to  that 
upon  BC ;  for  the  same  reason  as  DC 
to  CB,  so  is  the  figure  upon  CA  to  that 
upon  CB.  Wherefore,  as  BD  and  DC 
together  to  BC,  so  are  the  figures  upon 
BA  and  on  AC,  together,  to  the  figure 
upon  BC  (24.  5.) ;  therefore  the  figures  on  BA,  and  on  AC,  are  together 
equal  to  that  on  BC  ;  and  they  are  similar  figures. 

PROP.  XXXIL      THEOR. 

If  two  triangles,  which  have  two  sides  of  the  one  proportional  to  two  sides  oj 
the  other,  be  joined  at  one  angle ,  so  as  to  have  their  homologous  sides  pa- 
rallel to  one  another ;  their  remaining  sides  shall  be  in  a  straight  line. 

Let  ABC,  DCE  be  two  triangles  which  have  two  sides  BA,  AC  propor- 
tional to  the  two  CD,  DE,  viz.  BA  to  AC,  as  CD  to  DE  ;  and  let  AB  be 
parallel  to  DC,  and  AC  to  DE  ;  BC  and  CE  are  in  a  straight  line. 

Because  AB  is  parallel  to  DC,  and  the  straight  line  AC  meets  them,  the 
alternate  angles  BAG,  ACD  are  equal  (29  1.) ;  for  the  same  reason,  the 
angle  CDE  is  equal  to  the  angle 
ACD  ;  wherefore  also  BAC  is  equal 
to  CDE  :  And  because  the  triangles 
ABC,  DCE  have  one  angle  at  A 
equal  to  one  at  D,  aifl  the  sides  about 
these  angles  proportionals,  viz.  BA  to 
AC,  as  CD  to  DE,  the  triangle  ABC 
is  equiangular  (6.  6.)  to  DCE  : 
Therefore  the  angle  ABC  is  equal  to 


OF  GEOMETRY.    BOOK  VI. 


147 


the  angle  DCE  :  And  the  angle  BAG  was  proved  to  be  equal  to  ACD  : 
Therefore  the  whole  angle  ACE  is  equal  to  the  two  angles  ABC,  BAG  ; 
add  the  common  angle  ACB,  then  the  angles  ACE,  ACB  are  equal  to  the 
angles  ABC,  BAG,  ACB  :  But  ABC,  BAG,  ACB  are  equal  to  two  right 
angles  (32. 1.) ;  therefore  also  the  angles  AGE,  ACB  are  equal  to  two 
right  angles  :  And  since  at  the  point  C,  in  the  straight  line  AG,  the  two 
straight  lines  BC,  CE,  which  are  on  the  opposite  sides  of  it,  make  the  ad- 
jacent angles  ACE,  ACB  equal  to  two  right  angles  ;  therefore  (14. 1.)  BC 
and  CE  are  in  a  straight  line. 


PROP.  XXXIII.    THEOR. 


In  equal  circles,  angles,  whether  at  the  centres  or  circumferences,  have  the  same 
ratio  which  the  arcs,  on  which  they  stand,  have  to  one  another :  So  also  have 
the  sectors. 

Let  ABC,  DEF  be  equal  circles  ;  and  at  their  centres  the  angles  BGC, 
EHF,  and  the  angles  BAG,  EDF  at  their  circumferences  ;  as  the  arc  BC 
to  the  arc  EF,  so  is  the  angle  BGC  to  the  angle  EHF,  and  the  angle  BAG 
to  the  angle  EDF  :  and  also  the  sector  BGC  to  the  sector  EHF. 

Take  any  number  of  arcs  GK,  KL,  each  equal  to  BC,  and  any  number 
whatever  FM,  MN  each  equal  to  EF  ;  and  join  GK,  GL,  HM,  HN.  Be- 
cause the  arcs  BC,  GK,  KL  are  all  equal,  the  angles  BGC,  CGK,  KGL 
are  also  all  equal  (27.  3.) :  Therefore,  what  multiple  soever  the  arc  BL  is 
of  the  arc  BC,  the  same  multiple  is  the  angle  BGL  of  the  angle  BGC  :  For 
the  same  reason,  whatever  multiple  the  arc  EN  is  of  the  arc  EF  the  same 
multiple  is  the  angle  EHN  of  the  angle  EHF.  But  if  the  arc  BL,  be  equal 
to  the  arc  EN,  the  angle  BGL  is  also  equal  (27.  3.)  to  the  angle  EHN  ; 
or  if  the  arc  BL  be  greater  than  EN,  likewise  the  angle  BGL  is  greater 
than  EHN  :  and  if  less,  less  :  There  being  then  four  magnitudes,  the  two 
arcs,  BC,  EF,  and  the  two  angles  BGC,  EHF,  and  of  the  arc  BC,  and  of 
the  angle  BGC,  have  been  taken  any  equimultiples  whatever,  viz.  the  arc 
BL,  and  the  angle  BGL  ;  and  of  the  arc  EF,  and  of  the  angle  EHF,  any 
equimultiples  whatever,  viz.  the  arc  EN,  and  the  angle  EHN :  And  it 
has  been  proved,  that  if  the  arc  BL  be  greater  than  EN,  the  angle  BGL 
is  greater  than  EHN  ;  and  if  equal,  equal ;  and  if  less,  less  ;  As  therefore, 
the  arc  BC  to  the  arc  EF,  so  (def.  5. 5.)  is  the  angle  BGC  to  the  angle 


148  ELEMENTS 

EHF  :  But  as  the  angle  BGC  is  to  the  angle  EHF,  so  is  (15.  5.)  the  an- 
gle BAG  to  the  angle  EDF,  for  each  is  double  of  each  (20.  3.)  :  Therefore, 
as  the  circumference  BG  is  to  EF,  so  is  the  angle  BGC  to  the  angle  EHF, 
and  the  angle  BAG  to  the  angle  EDF. 

Also,  as  the  arc  BG  to  EF,  so  is  the  sector  BGG  to  the  sector  EHF. 
Join  BG,  GK,  and  in  the  arcs  BG,  GK  take  any  points  X,  0,  and  join  BX, 
XG,  GO,  OK  :  Then,  because  in  the  triangles  GBG,  GGK,  the  two  sides 
BG,  GG  are  equal  to  the  two  GG,  GK,  and  also  contain  equal  angles  ;  the 
base  BG  is  equal  (4.  1.)  to  the  base  GK,  and  the  triangle  GBG  to  the  tri- 
angle GGK  :  And  because  the  arc  BG  is  equal  to  the  arc  GK,  the  remain- 
ing part  of  the  whole  circumference  of  the  circle  ABG  is  equal  to  the  re- 
maining part  of  the  whole  circumference  of  the  same  circle  :  Wherefore 
the  angle  BXG  is  equal  to  the  angle  GOK  (27.  3.) ;  and  the  segment 
BXG  is  therefore  similar  to  the  segment  GOK  (def.  9.  3.) ;  and  they  are 
upon  equal  straight  lines  BG,  GK  :  But  similar  segments  of  circles  upon 
equal  straight  lines  are  equal  (24.  3.)  to  one  another  :  Therefore  the  seg- 
ment BXG  is  equal  to  the  segment  GOK  :  And  the  triangle  BGG  is  equal 
to  the  triangle  GGK  ;  therefore  the  whole,  the  sector  BGG  is  equal  to  the 
whole,  the  sector  GGK :  For  the  same  reason,  the  sector  KGL  is  equal  to 
each  of  the  sectors  BGG,  GGK  ;  and  in  the  same  manner,  the  sectors 
EHF,  FHM,  MHN,  may  be  proved  equal  to  one  another  :  Therefore,  what 
multiple  soever  the  arc  BL  is  of  the  arc  BG,  the  same  multiple  is  the  sec- 
tor BGL  of  the  sector  BGC.  For  the  same  reason,  whatever  multiple  the 
arc  EN  is  of  EF,  the  same  multiple  is  the  sector  EHN  of  the  sector  EHF  ; 
Now  if  the  arc  BL  be  equal  to  EN,  the  sector  BGL  is  equal  to  the  sector 


EHN  ;  and  if  the  arc  BL  be  greater  than  EN,  the  sector  BGL  is  greater 
than  the  sector  EHN  ;  and  if  less,  less  :  Since,  then,  there  are  four  mag- 
nitudes, the  two  arcs  BG,  EF,  and  the  two  sectors  BGG,  EHF,  and  of  the 
arc  BG,  and  sector  BGC,  the  arc  BL  and  the  sector  BGL  are  any  equi- 
multiples whatever ;  and  of  the  arc  EF,  and  sector  EHF,  the  arc  EN  and 
sector  EHN,  are  any  equimultiples  whatever  ;  and  it  has  been  proved,  that 
if  the  arc  BL  be  greater  than  EN,  the  sector  BGL  is  greater  than  the  sec- 
tor EHN  ;  if  equal,  equal;  and  if  less,  less  ;  therefore  (def.  5.  5.)  as  the 
arc  BC,  is  to  the  arc  EF,  so  is  the  sector  BGG  to  the  sector  EHF. 


1 


OF  GEOMETRY.    BOOK  VI. 


149 


PROP.  B.     THEOR. 

If  an  angle  of  a  triangle  he  bisected  hy  a  straight  line^  which  likewise  cuts  the 
base;  the  rectangle  contained  by  the  sides  of  the  triangle  is  equal  to  the 
rectangle  contained  by  the  segments  of  the  base,  together  with  the  square  of 
the  straight  line  bisecting  the  angle. 

Let  ABC  be  a  triangle,  and  let  the  angle  BAG  be  bisected  by  the 
straight  line  AD  ;  the  rectangle  BA.AC  is  equal  to  the  rectangle  BD.DC, 
together  with  the  square  of  AD. 

Describe  the  circle  (Prop.  5.  4.)  ACB  about 
the  triangle,  and  produce  AD  to  the  circum- 
ference in  E.  and  join  EC  Then,  because 
the  angle  BAD  is  equal  to  the  angle  CAE, 
and  the  angle  ABD  to  the  angle  (21.  3.) 
AEC,  for  they  are  in  the  same  segment ;  the 
triangles  ABD,  AEC  are  equiangular  to  one 
another :  Therefore  BA  :  AD  :  :  E A  :  (4.  6.) 

AC,  and  consequently,  BA.AC  =  (16.  6.) 
AD.AE=ED.DA  (3.  2.)  +DA2.  But  ED. 
DA=BD.DC,  therefore  BA.AC  =  BD.DC 
-f-DA2. 

PROP.  C.    THEOR. 

If  from  any  angUofa  triangle  a  straight  line  be  drawn  perpendicular  to  the 
base ;  the  rectangle  contained  by  the  sides  of  the  triangle  is  equal  to  the 
rectangle  contained  by  the  perpendicular,  and  the  diameter  of  the  circle  de- 
scribed about  the  triangle. 

Let  ABC  be  a  triangle,  and  AD  the  perpendicular  from  the  angle  A  to 
the  base  BC  ;  the  rectangle  BA.AC  is  equal  to  the  rectangle  contained  by 
AD  and  the  diameter  of  the  circle  described  about  the  triangle. 

Describe  (Prop.  5.  4.)  the  circle  ACB 
about  the  triangle,  and  draw  its  diameter 
AE,  and  join  EC ;  Because  the  right 
angle  BDA  is  equal  to  the  angle  ECA  in 
a  semicircle,  and  the  angle  ABD  to  the 
angle  AEC,  in  the  same  segment  (21. 
3.);  the  triangles  ABD,  AEC  are  equi- 
angular :    Therefore,  as  (4.  6.)  BA  to 

AD,  so  is  EA  to  AC  :  and  consequently 
the  rectangle  BA.AC  is  equal  (16.  6.)  to 
the  rectangle  EA.AD. 


IdO 


ELEMENTS 


PROP.  D.    THEOR. 


The  rectangle  contained  by  the  diagonals  of  a  quadrilateral  inscribed  in  a 
circle^  is  equal  to  both  the  rectangles^  contained  by  its  opposite  sides. 

Let  ABCD  be  any  quadrilateral  inscribed  in  a  circle,  and  let  AC,  BD  be 
drawn ;  the  rectangle  AC.BD  is  equal  to  the  two  rectangles  AB.CD,  and 
AD.BC. 

Make  the  angle  ABE  equal  to  the  angle  DBC  ;  add  to  each  of  these 
the  common  angle  EBD,  then  the  angle  ABD  is  equal  to  the  angle  EBC  : 
Antx  the  angle  BDA  is  equal  to  (21.  3.)  the  angle  BCE,  because  they  are 
in  the  same  segment ;  therefore  the  triangle 
ABD  is  equiangular  to  the  triangle  BCE. 
Wherefore  (4.  6.),  BC  :  CE  : :  BD  :  DA, 
and  consequently  (16.  6.)  BC.DA=BD.CE. 
Again,  because  the  angle  ABE  is  equal  to 
the  angle  DBC,  and  the  angle  (21.  3.)  BAE 
to  the  angle  BDC,  the  triangle  ABE  is  equi- 
angular to  the  triangle  BCD  ;  therefore  BA 
:  AE  : :  BD  :  DC,  and  BA.DC=BD.AE  : 
But  it  was  shewn  that  BC.DA=BD.CE ; 
wherefore  BC.DA  +  BA.DC  =  BD.CE+ 
BD.AE=BD.AC(1.2.).  That  is,  the  rect- 
angle contained  by  BD  and  AC,  is  equal  to  the  rectangles  contained  by 
AB,  CD,  and  AD,  BC. 


PROP.  E.    THEOR. 

If  an  arc  of  a  circle  be  bisected,  and  from  the  extremities  of  the  arc,  and  from 
the  point  of  bisection,  straight  lines  be  drawn  to  any  point  in  the  circum- 
ference, the  sum  of  the  two  lines  drawn  from  the  extremities  of  the  arc  will 
have  to  the  line  drawn  from  the  point  of  bisection,  the  same  ratio  which  the 
straight  line  subtending  the  arc  has  to  the  straight  line  subtending  half  the 
arc. 

Let  ABD  be  a  circle,  of  which  AB  is  an  arc  bisected  in  C,  and  from  A, 
C,  and  B  to  D,  any  point  whatever  in  the  circumference,  let  AD,  CD,  BD 
be  drawn  ;  the  sum  of  the  two  lines  AD 
and  DB  has  to  DC  the  same  ratio  that 
BA  has  to  AC. 

For  since  ACBD  is  a  quadrilateral  in- 
scribed in  a  circle,  of  which  the  diagonals 
are  AB  and  CD,  AD.CB+DB.AC  (D 
6.)  =  AB.CD  :  but  AD.CB+DB.AC  = 
AD.AC  -f-  DB.AC,  because  CB  =  AC. 
Therefore  AD.AC+DB.AC,  that  is  (1. 
2.),(AD+DB)  AC=AB.CD.  And  be- 
cause the  sides  of  equal  rectangles  are  re- 
ciprocally proportional  (14.  6.),  AD+DB 
.  DC  : :  AB  :  AC. 


OF  GEOMETRY.     BOOK  VI.  151 


PROP.  F.    THEOR. 


Jf  two  points  he  taken  in  the  diameter  of  a  circle,  such  that  the  rectangle  contained 
by  ihe  segments  intercepted  between  them  and  the  centre  of  the  circle  be  equal  to 
the  square  of  the  radius:  andiffrom  these  points  two  straight  lines  be  drawn 
to  any  point  whatsoever  in  the  circumference  of  the  circle,  the  ratio  of  these 
lines  will  be  the  same  with  the  ratio  of  the  segments  intercepted  between  the 
two  first  mentioned  points  and  the  circumference  of  the  circle. 

Let  ABC  be  a  circle,  of  which  the  centre  is  D,  and  in  DA  produced,  let 
the  points  E  and  F  be'  such  that  the  rectangle  ED,  DF  is  equal  to  the 
square  of  AD  ;  from  E  and  F  to  any  point  B  in  the  circumference,  let  EB, 
FB  be  drawn ;  FB  :  BE  :  :  FA  :  AE. 

Join  BD,  and  because  the  rectangle  FD,  DE  is  equal  to  the  square  of 
AD,  that  is,  of  DB,  FD  :  DB  :  :  DB  :  DE  (17.  6.). 

The  two  triangles,  FDB,  BDE  have  therefore  the  sides  proportional 
that  are  about  the  common  angle  D ;  therefore  they  are  equiangular  (6. 
6.),  the  angle  DEB  being,  equal  to  the  angle  DBF,  and  DBE  to  DFB. 


Now,  since  the  sides  about  these  equal  angles  are  also  proportional  (4.  6.), 
FB  :  BD  : :  BE  :  ED,  and  alternately  (16.  5:),  FB  :  BE  :  :  BD  :  ED,  or 
FB  :  BE  : :  AD  :  DE.  But  because  FD  :  DA  :  :  DA  :  DE,  by  division 
(17.  5.),  FA  :  DA  :  :  AE  :  ED,  and  alternately  (11.  5.),  FA  :  AE  : :  DA 
:  ED.  Now  it  has  been  shewn  that  FB  :  BE  :  :  AD  :  DE,  therefore  FB 
:  BE  :  :  FA  :  AE. 

Cor.  If  AB  be  drawn,  because  FB  :  BE  :  :  FA  :  AE,the  angle  FBE 
is  bisected  (3.  6.)  by  AB.  Also,  since  FD  :  DC  :  :  DC  :  DE,  by  compo- 
sition (18.  5.),  FC  :  DC  :  :  CE  :  ED,  and  since  it  has  been  shewn  that 
FA  :  AD  (DC)  :  :  AE  :  ED,  therefore,  ex  aquo,  FA  :  AE  :  :  FC  :  CE. 
But  FB  :  BE  : :  FA  :  AE,  therefore,  FB  :  BE  : :  FC  :  CE  (11. 5.),  so  that 
if  FB  be  produced  to  G,  and  if  BC  be  drawn,  the  angle  EBG  is  bisected 
by  the  line  BC  (A.  6.). 


152 


ELEMENTS 


PROP.  G.     THEOR. 

If  from  the  extremity  of  the  diameter  of  a  circle  a  straight  line  he  drawn  in  the 
circle,  and  if  either  within  the  circle  or  produced  without  it,  it  meet  a  line  per- 
pendicular  to  the  sdme  diameter,  the  rectangle  contained  hy  the  straight  line 
drawn  in  the  circle,  and  the  segment  nf  it,  intercepted  between  the  extremity 
of  the  diameter  and  the  perpendicular,  is  equal  to  the  rectangle  contained  hy 
the  diameter  and  the  segment  of  it  cut  off  by  the  perpendicular. 

Let  ABC  be  a  circle,  of  which  AC  is  a  diameter,  let  DE  be  perpendicu- 
lar to  the  diameter  AC,  and  let  AB  meet  DE  in  F  ;  the  rectangle  BA.AF 
is  equal  to  the  rectangle  CA.AD.     Join  BC,  and  because  ABC  is  an  an- 


gle in  a  semicircle,  it  is  a  right  angle  (31.  3.):  Now,  the  angle  ADF  is 
also  a  right  angle  (Hyp.) ;  and  the  angle  BAG  is  either  the  same  with 
DAF,  or  vertical  to  it ;  therefore  the  triangles  ABC,  ADF  are  equiangular, 
and  BA  :  AC  : :  AD  ;  AF  (4.  6.) ;  therefore  also  the  rectangle  BA.AF, 
contained  by  the  extremes,  is  equal  to  the  rectangle  ACAD  contained  by 
the  means  (16.  6.). 

PROP.  H.    THEOR. 


The  perpendiculars  drawn  from  the  three  angles  of  any  triangle  to  the  opposite 
sides  intersect  one  another  in  the  same  point. 

Let  ABC  be  a  triangle,  BD  and  CE  two  perpendiculars  intersecting  one 
another  in  F  ;  Let  AF  be  joined,  and  produced  if  necessary,  let  it  meet  BC 
in  G,  AG  is  perpendicular  to  BC. 

Join  DE,  and  about  the  triangleAEF  let  a  circle  be  described,  AEF  : 
then,  because  AEF  is  a  right  angle,  the  circle  described  about  the  triangle 
AEF  will  have  AF  for  its  diameter  (31.  3.).  In  the  same  manner,  the 
circle  described  about  the  triangle  ADF  has  AF  for  its  diameter ;  there- 
fore the  points  A,  E,  F  and  D,  are  in  the  circumference  of  the  same  circle. 


OF  GEOMETRY.    BOOK  VI. 


153 


But  because  the  angle  EFB  is  equal 
to  the  angle  DFC  (15. 1.),  and  also 
the  angle  BEF  to  the  angle  CDF, 
being  both  right  angles,  the  triangles 
BEF,  and  CDF  are  equiangular,  and 
therefore  BF  :  EF  : :  CF  :  FD  (4.  6.), 
or  alternately  (16. 5.)  BF  :  FC  :  :  EF 
:  FD.  Since,  then,  the  sides  about 
the  equal  angles  BFC,  EFD  are  pro- 
portionals, the  triangles  BFC,  EFD 
are  also  equiangular  (6.  6.) ;  where- 
fore the  angle  FCB  is  equal  to  the  an- 
gle EDF.  But  EDF  is  equal  to  EAF, 
because  they  are  angles  in  the  same 
segment  (21.  3.);  therefore  the  angle 
EAF  is  equal  to  the  angle  FCG  :  Now,  the  angles  AFE,  CFG  are  also 
equal,  because  they  are  vertical  angles  ;  therefore  the  remaining  angles 
AEF,  FGC  are  also  equal  (4.  Cor.'  32.  1.) :  But  AEF  is  a  right  angle, 
therefore  FGC  is  a  right  angle,  and  AG  is  perpendicular  to  BC. 

Cor.  The  triangle  ADE  is  similar  to  the  triangle  ABC.  For  the  two 
triangles  BAD,  CAE  having  the  angles  at  D  and  E  right  angles,  and  the 
angle  at  A  common,  are  equiangular,  and  therefore  B A  :  AD  : :  CA  :  AE, 
and  alternately  BA  :  CA  : :  AD  :  AE  ;  therefore  the  two  triangles  BAC, 
DAE,  have  the  angle  at  A  common,  and  the  sides  about  that  angle  pro- 
portionals, therefore  they  are  equiangular  (6.  6.)  and  similar. 

Hence  the  rectangles  BA.AE,  CA.AD  are  equal. 


PROP.  K.    THEOR. 


If  from  any  angle  of  a  triangle  a  perpendicular  he  drawn  to  the  opposite  side 
or  base :  the  rectangle  contained  by  the  sum  and  difference  of  the  other  two 
sides,  is  equal  to  the  rectangle  contained  by  the  sum  and  difference  of  the 
segments f  into  which  the  base  is  divided  by  the  perpendicular. 

Let  ABC  be  a  triangle,  AD  a  perpendicular  drawn  from  the  angle  A  on 
the  base  BC,  so  that  BD,  DC  are  the  segments  of  the  base  ;  (AC-f  AB) 
(AC-AB)=(CD+DB)  (CD-DB.) 


164 


ELEMENTS 


From  A  as  a  centre  with  the  radius  AC,  the  greater  of  the  two  sides, 
describe  the  circle  CFG  :  produce  AB  to  meet  the  circumference  in  E  and 
F,  and  CB  to  meet  it  in  G.  Then  because  AF=AC,  BF=AB+AC, 
the  sum  of  the  sides  ;  and  since  AE=AC,  BE=AC— AB=  the  diffe- 
rence of  the  sides.  Also,  because  AD  drawn  from  the  centre  cuts  GC  at 
right  angles,  it  bisects  it ;  therefore,  when  the  perpendicular  falls  within 
the  triangle,  BG=DG—DB=DC—DB=  the  difference  of  the  segments 
of  the  base,  and  BC=BD-}-DC=  the  sum  of  the  segments.  But  when 
AD  falls  without  the  triangle,  BG=DG4-DB=CD-f-DB=  the  sum  of 
the  segments  of  the  base,  and  BC=CD— DB=  the  difference  of  the  seg- 
ments of  the  base.  Now,  in  both  cases,  because  B  is  the  intersection  of 
the  two  lines  FE,  GC,  drawn  in  the  circle,  FB.BE=CB.BG  ;  that  is,  as 
has  been  shewn,  (xlC  +  AB)  (AC-AB)=(CD4-DB)  (CD-DB). 


PROBLEMS 

RELATING  TO  THE  SIXTH  BOOK. 


PROP.L.    PROBI^EM. 
To  construct  a  square  that  shall  he  equivalent  to  a  given  fectilineal figure.  ■ 

Let  A  be  the  given  rectilineal  figure  ;  it  is  required  to  describe  a  square 
that  shall  be  equivalent  to  A. 

Describe  (Prop.  45.1.)  the 
rectangular  parallelogram 
BCDE  equivalent  to  the  rec- 
tilineal figure  A ;  produce 
one  of  the  sides  BE,  of  this 
rectangle,  and  make  EF  = 
ED ;  bisect  BF  in  G,  and 
from  the  centre  G,  at  the 
distance  GB,  or  GF,  de- 
scribe the  semicircle  BHF, 
and  produce  DE  to  H. 

HE^==BE  X  EF,  (13.  6.) ;  therefore  the  square  described  upon  HE  will 
be  equivalent  to  the  rectilineal  figure  A. 

,  SCHOLIUM. 

This  problem  may  be  considered  as  relating  to  the  second  Book  :  Thus, 
join  GH,  the  rest  of  the  construction  being  the  same,  as  above  ;  because 
the  straight  line  BF  is  divided  into  two  equal  parts  in  the  point  G,  and  into 
two  unequal  in  the  point  E,  the  rectangle  BE.EF,  together  with  the  square 
of  EG,  is  equal  (5.  2.)  to  the  square  of  GF  :  but  GF  is  equal  to  GH  ; 


OF  GEOMETRY.    BOOK  VI. 


165 


therefore  the  rectangle  BE,  EF,  together  with  the  square  of  EG,  is  equal 
to  the  square  of  G^  :  But  the  squares  of  HE  and  EG,  are  equal  (47.  1.) 
to  the  square  of  GH  :  Therefore  also  the  rectangle  BE.EF,  together  with 
the  square  of  EG,  is  equal  to  the  squares  of  HE  and  EG.  Take  away 
the  square  of  EG,  which  is  common  to  both,  and  the  remaining  rectangle 
BE.EF  is  equal  to  the  square  of  EH  :  But  BD  is  the  rectangle  contained 
by  BE  and  EF,  because  EF  is  equal  to  ED  ;  therefore  BD  is  equal  to  the 
square  of  EH  ;  and  BD  is  also  equal  to  the  rectilineal  figure  A  ;  therefore 
the  rectilineal  figure  A  is  equal  to  the  square  of  EH  :  Wherefore  a  square 
has  been  made  equal  to  the  given  rectilineal  figure  A,  viz.  the  square  de- 
scribed upon  EH. 

Note.   This  operation  is  called  squaring  the  rectilineal  figure,  or  finding 
the  quadrature  of  it. 

PROP.  M.     PROB. 


To  construct  a  rectangle  that  shall  he  equivalent  to  a  given  square,  and  the 
difference  of  whose  adjacent  sides  shall  he  equal  to  a  given  line. 

Suppose  C  equal  to  the  given  square,  and 
AB  the  difference  of  the  sides. 

Upon  the  given  line  AB  as  a  diameter,  de- 
scribe a  circle  ;  at  the  extremity  of  the  diam- 
eter draw  the  tangent  AD  equal  to  the  side 
of  the  square  C  ;  through  the  point  D,  and  the 
centre  O,  draw  the  secant  DF  ;  then  will  DE 
and  DF  be  the  adjacent  sides  of  the  rectangle 
required. 

First,  the  difference  of  their  sides  is  equal 
to  the  diameter  EF  or  AB  ;  secondly,  the  rect- 
angle DE.DF  is  equal  to  AD^  (36.  3.) ;  hence 
that  rectangle  is  equivalent  to  the  given  square  C. 


PROP.  N.      PROB. 


To  construct  a  rectangle  equivalent  to  a  given  square,  and  having  the  sum 
of  its  adjacent  sides  equal  to  a  given  line. 

Let  C  be  the  given  square,  and  AB  equal  to  the  sum  of  the  sides  of  the 
required  triangle. 

Upon  AB  as  a  diameter, 
describe  a  semicircle  ;  draw 
the  line  DE  parallel  to  the 
diameter,  at  a  distance  AD 
from  it,  equal  to  the  side  of 
the  given  square  C ;  from  the 

point  E,  where  the  parallel      -^  F   B 

cuts  the  circumference,  draw  EF  perpendicular  to  the   diameter ;  AF 
and  FB  will  be  the  sides  of  the  rectangle  required. 


156 


ELEMENTS 


For  their  sum  is  equal  to  AB  ;  and  their  rectangle  AF.FB  is  equal  to  the 
square  EF,  or  to  the  square  AD  ;  hence  that  rectangle  is  equivalent  to  the 
given  square  C. 

SCHOLIUM. 

To  render  the  problem  possible,  the  distance  AD  must  not  exceed  the 
radius  ;  that  is,  the  side  of  the  square  C  must  not  exceed  the  half  of  the 
line  AB. 

PROP.    O.     PROB. 


To  construct  a  square  that  shall  he  to  a  given  square  as  a  given  line  to  a  given 

line. 

Upon  the  indefinite  straight  line  GH  take  GK=E,  and  KH=F ;  de- 
scribe on  GH  a  semicircle,  and  draw  the  perpendicular  KL.  Through 
the  points  G,  H,  draw  the 
straight  lines  LM,  LN,  mak- 
ing the  former  equal  AB,  the 
side  of  the  given  square,  and 
through  the  point  M,  draw 
MN  parallel  to  GH,  then  will 
LN  be  the  side  of  the  square 
sought. 

For,  since  MN  is  parallel      . —       -rvr  -kt 

to  GH,  LM  :   LN  :  :  LG  :  M  N 

LH  ;  consequently,  LM*  :  LN*  : :  LG*  ;  LH*  (22.  6.) ;  but,  since  the  trian- 
gle LGH  is  right  angled,  we  have  LG*  :  LH*  ;  :  GK  :  KH  ;  hence  LM*  : 
LN*  :  :  GK  :  KH  ;  but,  by  construction  GK=E,  and  KH=F,  also  LM 
=AB  ;  therefore,  the  square  described  on  AB  is  to  that  described  on  LN, 
as  the  line  E  is  to  the  line  F. 


PROP.  P.    PROB. 

To  divide  a  triangle  into  two  parts  by  a  line  from  the  vertex  of  one  of  its  angles, 
so  that  the  parts  may  be  to  each  other  as  a  straight  line  M  to  another  straight 
line  N. 

Divide  BC  into  parts  BD,  DC  propor- 
tional to  M,  N ;  draw  the  line  AD,  and 
the  triangle  x^BC  will  be  divided  as  re- 
quired. 

For,  since  the  triangles  of  the  same 
altitude  are  to  each  other  as  their  bases, 
we  have  ABD  :  ADC  : :  BD  :  DC  :  : 
M:  N. 

SCHOLIUM. 

A  triangle  may  evidently  be  divided  into  any  number  of  parts  propor- 
tional to  given  lines,  by  dividing  the  base  in  the  same  proportion. 


OF  GEOMETRY.  BOOK  VI.  157 


PROP.  Q.  PROB. 

To  divide  a  triangle  into  two  parts  hy  a  tine  drawn  parallel  to  one  of  its  sides, 
so  that  these  parts  may  he  to  each  other  as  two  straight  lines  M,  N. 

As  M+N  :  N,  so  make  AB^  to  AD^ 
(Prob.  4.) ;  Draw  DE  parallel  to  BC, 
and  the  triangle  is  divided  as  required. 

For  the  triangles  ABC,  ADE  being 
similar,  ABC  :  ADE  : :  AB2 :  AD2  ;  but 
M+N :  N : :  AB^ :  AD^ ;  therefore  ABC 
:  ADE  :  :  M-f-N  :  N;  consequently 
BDEC  ;  ADE  : :  M  :  N. 


PROP.    R.      PROB. 

To  divide  a  triangle  into  two  partSj  hy  a  line  drawn  from  a  given  point  tn 
one  of  its  sides ^  so  that  the  parts  may  he  to  each  other  as  two  given  lines 

M,  N. 

Let  ABC  be  the  given  triangle,  and  P  the  given  point ;  draw  PC,  and. 
divide  AB  in  D,  so  that  AD  is  to  DB  as  M  is  to  N ;  draw  DE  parallel  to 
PC,  join  PE,  and  the  triangle  will  be  divid- 
ed by  the  line  PE  into  the  proposed  parts. 

For  join  DC ;  then  because  PC,  DE  are 
parallel,  the  triangles  PDE,  CDE  are  equal ; 
to  each  add  the  triangle  DEB,  then  PEB=: 
DCB  ;  and  consequently,  by  taking  each  from 
the  triangle  ABC,  there  results  the  quadri- 
lateral ACEP  equivalent  to  the  triangle  .^ 
ACD.  B 

Now,  ACD  :  DCB  : :  AD  :  DB  : :  M  :  N ;  consequently, 
ACEP  :  PEB  :  :  M  :  N 

SCHOLIUM. 

The  above  operation  suggests  the  method  of  dividing  a  triangle  into  any 
number  of  equal  parts  by  lines  drawn  from  a  given  point  in  one  of  its  sides ; 
for  if  AB  be  divided  into  equal  parts,  and  lines  be  drawn  from  the  points  of 
equal  division,  parallel  to  PC,  they  will  intersect  BC,  and  AC^  and  from 
these  several  points  of  intersection  if  lines  be  drawn  to  P,  they  will  divide 
the  triangle  into  equal  parts. 


158  ELEMENTS 


PROP.    S.        PROB. 


To  divide  a  triangle  into  three  equivalent  parts  hy  lines  dravm  from  the  vcr- 
tices  of  the  angles  to  the  same  point  within  the  triangle. 

Make  BD  equal  to  a  third  part  of  BC,  and  draw  DE  parallel  to  BA,  the 
side  to  which  BD  is  adjacent.  From  F,  the  middle  of  DE,  draw  the 
straight  lines  FA,  FB,  FC,  and  they  will 
divide  the  triangle  as  required. 

For,  draw  DA  ;  then  since  BD  is  one 
third  of  BC,  the  triangle  ABD  is  one 
third  of  the  triangle  ABC  ;  but  ABD= 
ABF  (37.  1.)  ;  therefore  ABF  is  one 
third  of  ABC  ;  also,  since  DF=FE, 
BDF  =  AFE  ;  likewise  CFD  =  CFE  , 
consequently  the  whole  triangle  FBC 
is  equal  to  the  whole  triangle  FCA ;  and 
FBA  has  been  shown  to  be  equal  to  a  third  part  of  the  whole  triangle 
ABC  ;  consequently  the  triangles  FBA,  FBC,  FCA,  are  each  equal  to  a 
third  part  of  ABC. 

PROP.  T.      PROB. 

To  divide  a  triangle  into  three  equivalent  parts ,  hy  lines  drawn  Jrom  a  given 

point  within  it. 

Divide  BC  into  three  equal  parts  in  the  points  D,  E,  and  draw  PD,  PE  ; 
draw  also  AF  parallel  to  PD,  and  AG  parallel  to  PE;  then  if  the  Imes 
PF,  PG,  PA  be  drawn,  the  trian-  « 

gle  ABC  will  be  divided  by  them 
into  three  equivalent  parts. 

For,  join  AD,  AE  ;  then  because 
AF,  PD  are  parallel,  the  triangle 
AFP  is  equivalent  to  the  triangle 
AFD  ;  consequently,  if  to  each  of 
these  there  be  added  the  triangle 
ABF,  there  will  result  the  quadri- 
lateral ABFP  equivalent  to  the 
triangle  ABD  ;  but  since  BD  is  a 
third  part  of  BC,  the  triangle  ABD 
is  a  third  part  of  the  triangle  ABC  ; 
consequently  the  quadrilateral  ABFP  is  a  third  part  of  the  triangle  ABC. 
Again,  because  AG,  PE  are  parallel,  the  triangle  AGP  is  equivalent  to 
the  triangle  AGE  and  if  to  each  of  these  there  be  added  the  triangle  ACG 
the  quadrilateral  ACGP  will  be  equivalent  to  the  triangle  ACE  ;  but  this 
triangle  is  one  third  of  ABC  ;  hence  the  quadrilateral  ACGP  is  one  third 
of  the  triangle  ABC :  consequently,  the  spaces  ABFP,  ACPG,  PFG  are 
each  equal  to  a  third  part  of  the  triangle  ABC. 


OF  GEOMETRY.    BOOK  VI 


159 


PROP.  U.     PROB. 


To  divide  a  quadrilateral  into  two  parts  hy  a  straight  line  drawn  from  the  vertex 
of  one  of  its  angles^  so  that  the  parts  may  he  to  each  other  as  a  line  M  to  an- 
other line  N. 

Draw  CE  perpendicular  to  AB,  and  construct  a  rectangle  equivalent  to 
the  given  quadrilateral,  of  which  one  side  maybe  CE  ;  let  the  other  side 
be  EF ;  and  divide  EF  in  G,  so  that 
M  :  N  :  :  GF  :  EG  ;  take  BP  equal 
to  twice  EG,  and  join  PC,  then  the 
quadrilateral  will  be  divided  as  re- 
quired.       4 

For,  by  construction,  the  triangle 
CPB  is  equivalent  to  the  rectangle 
CE.EG ;  therefore  the  rectangle  CE, 
GF  is  to  the  triangle  CPB  as  GF  is 
to  EG.  Now  CE.GF  is  equivalent 
to  the  quadrilateral  DP,  and  GF  is  to  EG  as  M  is  to  N  ;  therefore, 

DP  :  CPB  : :  M  :  N  ; 
that  is,  the  quadrilateral  is  divided,  as  required. 


PROP.  W.     PROB. 

To  divide  a  quadrilateral  into  two  parts  by  a  line  parallel  to  one  of  its  sides f 
so  that  these  parts  may  be  to  each  other  as  the  line  M  is  to  the  line  N. 

Produce  AD,  BC  till  they  meet  in  E  ;  draw  the  perpendicular  EF  and 
bisect  it  in  G.  Upon  the  side  GF  construct  a  rectangle  equivalent  to  the 
triangle  EDC,  and  let  HB  be  equal 
to  the  other  side  of  this  rectangle. 
Divide  AH  in  K,  so  that  AK  :  KH 
:  :  M  :  N,  and  as  AB  is  to  KB,  so 
make  EA^  to  Ea^ ;  draw  ab  paral- 
lel to  AB,  and  it  will  divide  the  quad- 
rilateral into  the  required  parts. 

For  since  the  triangles  EAB,  Eab 
are  similar,  we  have  the  proportion 

EAB  :Eab::  EA^  :  Ea^;    but  by        .       __    ^        ^^   ^ 

construction,  EA2   :   Ea2   :  :    AB   :       -^  iL.    M        H   i> 

KB  ;  so  that  EAB  :  Eab  :  :  AB  :  KB  :  :  AB.GF  :  KB.GF  ;  and  conse- 
quently, since  by  construction  EAB==AB.GF,  it  follows  that  Eab=KB. 
GF,  and  therefore  AK.GF=A5,  and  since  by  construction  AH. GF= AC, 
it  follows  that  KH.GF=aC.  Now  AK.GF  :  KH.GF  :  :  AK  :  KH  ;  but 
AK :  KH  :  :  M  :  N  ;  consequently, 

Ai:aC::M:N; 
that  is,  the  quadrilateral  is  divided,  as  required. 


160 


ELEMENTS,  &c. 


PROP.  X.   PROB. 

To  divide  a  quadrilateral  into  two  parts  by  a  line  drawn  from  a  point  in  one  of 
its  sides,  so  that  the  parts  may  he  to  each  other  as  a  line  M  is  to  a  line  N, 

Draw  PD,  upon  which  construct  a  rectangle  equivalent  to  the  given 
quadrilateral,  and  let  DK  be  the  other 
side  of  this  rectangle  ;  divide  DK  in 
L,  so  that  DL  :  LK  :  :  M  :  N  ;  make 
DF=2DL,  and  FG  equal  to  the  per- 
pendicular A.a ;  draw  Gp  parallel  to 
DP  ;  join  the  points  P,  p,  and  the 
quadrilateral  figure  will  be  divided, 
as  required. 

For  draw  the  perpendicular  ph  ; 
then  by  construction,  PD.DK  =  AC, 
and  PD.DF  =  PD.Aa  +  VDph,  that 
is,  PD.DF  is  equivalent  to  twice  the 
sum  of  the  triangles  APD,  pVD , 
consequently,  since  DL  is  half  DF, 
PD.DL=AP;7D  ;  and  therefore  PD. 
LK=PBC;> ;  but  PD.DL  :  PD.LK  :  :  DL  :  LK  : :  M  :  N ;  consequently, 

AP^D  :  PBC;» :  :  M  :  N  ; 
hence  the  quadrilateral  is  divided,  as  required. 


PROP.  Y.        PROB. 


To  divide  a  quadrilateral  hy  a  line  perpendicular  to  one  of  its  sides,  so  that  the 
two  parts  may  be  to  each  other  as  a  Une  M  is  to  a  line  N. 

Let  ABCD  be  the  given  quadrilateral,  which  is  to  be  divided  in  the  ratio 
of  M  to  N  by  a  perpendicular  to  the  side  AB. 

Construct  on  DE  perpendicular 
to  AB,  a  rectangle  DE.EF,  equi- 
valent to  the  quadrilateral  AC, 
and  divide  FE  in  G,  so  that  FG  : 
GE  :  :  M  :  N.  Bisect  AE  in  H, 
and  divide  the  quadrilateral  EC 
into  two  parts  by  a  line  PQ,  paral- 
lel to  DE,  so  that  those  parts  may 
be  to  each  other  as  FG  is  to  GH, 
then  PQ  will  also  divide  the  quadri- 
lateral AC  as  required. 

For,  by  construction  DE.EF=AC,  and  DE.EH=DAE  ;  hence  DE. 
HF=EC,  and  consequently,  since  the  quadrilateral  EC  is  divided  in  the 
same  proportion  as  the  base  FH  of  its  equivalent  rectangle,  it  follows  that 
QC=DE.FG,  and  EP=DE.GH,  also  AE==DE.GE  ;  consequently, 

QC  :  AP  :  :  FG ;  GE  :  :  M  :  N ; 
that  is,  the  quadrilateral  is  divided,  as  required. 


FoaThce 


SUPPLEMENT 

TO  THE 

ELEMENTS 

OP 

GEOMETRY. 


21 


ELEMENTS 


OF 


GEOMETRY. 


SUPPLEMENT. 


BOOK  I. 

OF  THE  QUADRATURE  OF  THE  CIRCLE, 


LEMMA 

Any  curve  line,  or  any  polygonal  line,  which  envelopes  a  convex  line  from  one 
end  to  the  other,  is  longer  than  the  enveloped^ line. 

Let  AMB  be  the  enveloped  line ;  then  will  it  be  less  tlian  the  lino 
APDB  which  envelopes  it. 

We  have  already  said  that  by  the 
term  convex  line  we  understand  a  line, 
polygonal,  or  curve,  or  partly  curve  and 
partly  polygonal,  such  that  a  straight 
line  cannot  cut  it  in  more  than  two 
points.  If  in  the  line  AMB  there  were 
any  sinuosities  or  re-entrant  portions,  it 
would  cease  to  be  convex,  because  a 
straight  line  might  cut  it  in  more  than 
two  points.  The  arcs  of  a  circle  are  essentially  convex ;  but  the  present 
proposition  extends  to  any  line  which  fulfils  the  required  conditions. 

This  being  premised,  if  the  line  AMB  is  not  shorter  than  any  of  those 
which  envelope  it,  there  will  be  found  among  the  latter,  a  line  shorter  than 
all  the  rest,  which  is  shorter  than  AMB,  or,  at  most,  equal  to  it.  Let 
ACDEB  be  this  enveloping  line :  any  where  between  those  two  lines, 
draw  the  straight  line  PQ,  not  meeting,  or  at  least  only  touching,  the  line 
AMB.  The  straight  line  PQ  is  shorter  than  PCDEQ  ;  hence,  if  instead 
of  the  part  PCDEQ,  we  substitute  the  straight  line  PQ,  the  enveloping  line 
APQB  will  be  shorter  than  APDQB.  But,  by  hypothesis,  this  latter  was 
shorter  than  any  other ;  hence  that  hypothesis  was  false  ;  hence  all  of  the 
enveloping  lines  are  longer  than  AMB 


F 


164  SUPPLEMENT  TO  THE  ELEMENTS 

Cor.  1.  Hence  the  perimeter  of  any  polygon  inscribed  in  a  circle  is 
less  than  the  circumference  of  the  circle. 

Cor.  2.  If  from  a  point  two  straight  lines  be  drawn,  touching  a  circle, 
these  two  lines  are  together  greater  than  the  arc  intercepted  between 
them ;  and  hence  the  perimeter  of  any  polygon  described  about  a  circle  is 
greater  than  the  circumference  of  the  circle. 

PROP.  L     THEOR. 

Ifjrom  the  greater  of  two  unequal  magnitudes  there  he  taken  away  its  half, 
and  from  the  remainder  its  half;  and  so  on  ;  There  will  at  length  remain 
a  magnitude  less  than  the  least  of  the  proposed  magnitudes. 

Let  AB  and  C  be  two  unequal  magnitudes,  of  which  AB  is  the  greater. 
If  from  AB  there  be  taken  away  its  half,  and  from  the 
remainder  its  half,  and  so  on ;  there  shall  at  length 
remain  a  magnitude  less  than  C.  * 

For  C  may  be  multiplied  so  as,  at  length,  to  be- 
come greater  than  AB.     Let    DE,  therefore,  be   a     -j^i 
multiple  of  C,  which  is  greater  than  AB,  and  let  it     ■    "^ 
contain  the   parts   DF,  FG,   GE,  each  equal  to  C. 
From  AB  take  BH  equal  to  its  half;  and  from  the      tt. 
remainder  AH,  take  HK  equal  to  its  half,  and  so  on, 
imtil  there  be  as  many  divisions  in  AB  as  there  are 
in  DE  ;  And  let -the  divisions  in  AB  be  AK,  KH, 
HB.     And  because  DE  is  greater  than  AB,  and  EG 
taken  from  DE  is  not  greater  than  its  half,  but  BH 
taken  from  AB  is  equal  to  its  half;  therefore  the  re- 
mainder GD    is   greater    than    the   remainder   HA.       S      C      H 
Again,  because  GD  is  greater  than  HA,  and  GF  is 
not  greater  than  the  half  of  GD,  but  HK  is  equal  to  the  half  of  HA ;  there- 
fore the  remainder  FD  is  greater  than  the  remainder  AK.      And  FD  is 
equal  to  C,  therefore  C  is  greater  than  AK  ;  that  is,  AK  is  less  than  C. 

PROP.  11.    THEOR. 

Equilateral  polygons^  of  the  same  number  of  sides,  inscribed  in  circles,  are 
similar,  and  are  to  one  another  as  the  squares  of  the  diameters  of  the 
circles. 

Let  ABCDEF  and  GHIKLM  be  two  equilateral  polygons  of  the  same 
number  of  sides  inscribed  in  the  circles  ABD  and  GHK  ;  ABCDEF  and 
GHIKLM  are  similar,  and  are  to  one  another  as  the  squares  of  the  diame- 
ters of  the  circles  ABD,  GHK. 

Find  N  and  0  the  centres  of  the  circles,  join  AN  and  BN,  as  also  GO 
and  HO,  and  produce  AN  and  G©  till  they  meet  the  circumferences  in  D 
andK. 

Because  the  straight  lines  AB,  BC,  CD,  DE,  EF,  FA,  are  all  equal, 
the  arcs  AB,  BC,  CD,  DE,  EF,  FA  are  also  equal  (28.  3.).  For  the 
same  reason,  the  arcs  GH,  HI,  IK,  KL,  LM,  MG  are  all  equal,  and  ihey 


OF  GEOxMETRY.    BOOK  I. 
B..- -.C  31 


165 


are  equal  in  number  to  the  others  ;  therefore,  whatever  part  the  arc  AB  is 
of  the  whole  circumference  ABD,  the  same  is  the  arc  GH  of  the  circum- 
ference GHK.  But  the  angle  ANB  is  the  same  part  of  four  right  angles, 
that  the  arc  AB  is  of  the  circumference  ABD  (33.  6.) ;  and  the  angle 
GOH  is  the  same  part  of  four  right  angles,  that  the  arc  GH  is  of  the  cir- 
cumference GHK  (33.  6.),  therefore  the  angles  ANB,  GOH  are  each  of 
them  the  same  part  of  four  right  angles,  and  therefore  they  are  equal  to 
one  another.  The  isosceles  triangles  ANB,  GOH  are  therefore  equian- 
gular, and  the  angle  ABN  equal  to  the  angle  GHO  ;  in  the  same  manner, 
by  joining  NO,  01,  it  may  be  proved  that  the  angles  NBC,  OHI  are  equal 
to  one  another,  and  to  the  angle  ABN.     Therefore  the  whole  angle  ABC 


is  equal  to  the  whole  GHI ;  and  the  same  may  be  proved  of  the  angles 
BCD,  HHC,  and  of  the  rest.  Therefore,  the  polygons  ABCDEF  and 
GHIKLM  are  equiangular  to  one  another  ;  and  since  they  are  equilateral, 
the  sides  about  the  equal  angles  are  proportionals  ;  the  polygon  ABCDEF 
is  therefore  similar  to  the  polygon  GHIKLM  (def.  1.6.).  And  because  simi- 
lar polygons  are  as  the  squares  of  their  homologous  sides  (20.  6.),  the  po- 
lygon ABCDEF  is  to  the  polygon  GHIKLM  as  the  square  of  AB  to  the 
square  of  GH  ;  but  because  the  triangles  ANB,  GOH  are  equiangular, 
the  square  of  AB  is  to  the  square  of  GH  as  the  square  of  AN  to  the  square 
of  GO  (4.  6.),  or  as  four  times  the  square  of  AN  to  four  times  the  square 
(15.  5.)  of  GO,  that  is,  as  the  square  of  AD  to  the  square  of  GK,  (2.  Cor. 
8. 2.).     Therefore  also,  the  polygon  ABCDEF  is  to  the  polygon  GHIKLM 


166  SUPPLEMENT  TO  THE  ELEMENTS 

as  the  square  of  AD  to  tlie  square  of  GK ;  and  they  have  also  been  shewn 
to  be  similar. 

CoR.  Every  equilateral  polygon  inscribed  in  a  circle  is  also  equiangu 
lar  :  For  the  isosceles  triangles,  which  have  their  common  vertex  in  the 
centre,  are  all  equal  and  similar ;  therefore,  the  angles  at  their  bases  are 
all  equal,  and  the  angles  of  the  polygon  are  therefore  also  equal. 

PROP.  in.     PROB. 

The  side  of  any  equilateral  polygon  inscribed  in  a  circle  heing  given,  to  find  the 
side  of  a  polygon  of  the  same  number  of  sides  described  about  the  circle. 

Let  ABCDEF  be  an  equilateral  polygon  inscribed  in  the  circle  ABD  ; 
it  is  required  to  find  the  side  of  an  equilateral  polygon  of  the  same  number 
of  sides  described  about  the  circle. 

Find  G  the  centre  of  the  circle  ;  join  GA,  GB,  bisect  the  arc  AB  in  H ; 
and  through  H  draw  KHL  touching  the  circle  in  H,  and  meeting  GA  and 
GB  produced  in  K  and  L  ;  KL  is  the  side  of  the  polygon  required. 

Produce  GF  to  N,  so  that  GN  maybe  equal  to  GL  ;  join  KN,  and  from 
G  draw  GM  at  right  angles  to  KN,  join  also  HG. 

Because  the  arc  AB  is  bisected  in  H,  the  angle  AGH  is  equal  to  the 
angle  BGH  (27.  3.) ;  and  because 
KL  touches  the  circle  in  H,  the 
angles  LHG,  KHG  are  right  an- 
gles (18.  3.);  therefore,  there  are 
two  angles  of  the  triangle  HGK, 
equal  to  two  angles  of  the  triangle 
HGL,  each  to  each.  But  the  side 
GH  is  common  to  these  triangles  ; 
therefore  they  are  equal  (26. 1.), and 
GL  is  equal  to  GK.  Again,  in 
the  triangles  KGL,  KGN,  because 
GN  is  equal  to  GL  ;  and  GK  com- 
mon, and  also  the  angle  LGK  equal 
to  the  angle  KGN  ;  therefore  the 
base  KL  is  equal  to  the  base  KN 
(4.  1.).  But  because  the  triangle  KGN  is  isosceles,  the  angle  GKN  is 
equal  to  the  angle  GNK,  and  the  angles  GMK,  GMN  are  both  right  an- 
gles by  construction ;  wherefore,  the  triangles  GMK,  GMN  have  two  an- 
gles of  the  one  equal  to  two  angles  of  the  other,  and  they  have  also  the 
side  GM  common,  therefore  they  are  equal  (26.  l.),and  the  side  KM  is  equal 
to  the  side  MN,  so  that  KN  is  bisected  in  M.  But  KN  is  equal  to  KL, 
and  therefore  their  halves  KM  and  KH  are  also  equal.  Wherefore,  in  the 
triangles  GKH,  GKM,  the  two  sides  GK  and  KH  are  equal  to  the  two 
GK  and  KM,  each  to  each ;  and  the  angles  GKH,  GKM,  are  also  equal, 
therefore  GM  is  equal  to  GH  (4.  1.) ;  wherefore,  the  point  M  is  in  the  cir- 
cumference of  the  circle  ;  and  because  KMG  is  a  right  angle,  KM  touches 
the  circle.  And  in  the  same  manner,  by  joining  the  centre  and  the  other 
angular  points  of  the  inscribed  polygon,  an  equilateral  polygon  may  bo 


OF  GEOMETRY.     BOOK  I.  167 

described  about  the  circle,  the  sides  of  which  will  each  be  equal  to  KL,  and 
will  be  equal  in  number  to  the  sides  of  the  inscribed  polygon.  Therefore, 
KL  is  the  side  of  an  equilateral  polygon,  described  about  the  circle,  of  the 
same  number  of  sides  with  the  inscribed  polygon  ABCDEF. 

Cor.  1.  Because  GL,  GK,  GN,  and  the  other  straight  lines  drawn 
from  the  centre  G  to  the  angular  points  of  the  polygon  described  about  the 
circle  ABD  are  all  equal ;  if  a  circle  be  described  from  the  centre  G,  with 
the  distance  GK,  the  polygon  will  be  inscribed  in  that  circle  ;  and  there- 
fore it  is  similar  to  the  polygon  ABCDEF. 

Cor.  2.  It  is  evident  that  AB,  a  side  of  the  inscribed  polygon,  is  to  KL, 
a  side  of  the  circumscribed,  as  the  perpendicular  from  G  upon  AB,  to  the 
perpendicular  from  G  upon  KL,  that  is,  to  the  radius  of  the  circle  ;  there- 
fore also,  because  magnitudes  have  the  same  ratio  with  their  equimultiples 
(15.  5.),  the  perimeter  of  the  inscribed  polygon  is  to  the  perimeter  of  the 
circumscribed,  as  the  perpendicular  from  the  centre,  on  a  side  of  the  in- 
scribed polygon,  to  the  radius  of  the  circle. 

PROP.  IV.    THEOR. 

A  circleheinggwen,two  similar  polygons  may  he  found,  the  one  described  about 
the  circle  y  and  the  other  inscribed  in  it,  which  shall  differ  from  one  another  by 
a  space  less  than  any  given  space. 

Let  ABC  be  the  given  circle,  and  the  square  of  D  any  given  space  ;  a 
polygon  may  be  inscribed  in  the  circle  ABC,  and  a  similar  polygon  describ- 
ed about  it,  so  that  the  difference  between  them  shall  be  less  than  the 
square  of  D. 

In  the  circle  ABC  apply  the  straight  line  AE  equal  to  D,  and  let  AB  be 
a  fourth  part  of  the  circumference  of  the  circle.  From  the  circumference 
AB  take  away  its  half,  and  from  the  remainder  its  half,  and  so  on  till  the 
circumference  AF  is  found  less  tian  the  circumference  AE  (1.1.  Sup.). 
Find  the  centre  G  ;  draw  the  diameter  AC,  as  also  the  straight  lines  AF 
and  FG  ;  and  having  bisected  the  circumference  AF  in  K,  join  KG,  and 
draw  HL  touching  the  circle  in  K,  and  meeting  GA  and  GF  produced  in 
H  and  L  ;  join  CF. 

Because  the  isosceles  triangles  HGL  and  AGF  have  the  common  an- 
gle AGF,  they  are  equiangular  (6.  6.)  and  the  angles  GHK,  GAF  are 
therefore  equal  to  one  another.  But  the  angle  GKH,  CFA  are  also  equal, 
for  they  are  right  angles ;  therefore  the  triangles  HGK,  ACF,  are  like- 
wise equiangular  (4.  Cor.  32. 1.). 

And  because  the  arc  AF  was  found  by  taking  from  the  arc  AB  its  half, 
and  from  that  remainder  its  half,  and  so  on,  AF  will  be  contained  a  certain 
number  of  times,  exactly,  in  the  arc  AB,  and  therefore  it  will  also  be  con- 
tained a  certain  number  of  times,  exactly,  in  the  whole  circumference 
ABC  ;  and  the  straight  line  AF  is  therefore  the  side  of  an  equilateral  poly- 
gon inscribed  in  the  circle  ABC.  Wherefore  also,  HL  is  the  side  of  an 
equilateral  polygon,  of  the  same  number  of  sides,  described  about  ABC  (3. 
1.  Sup.).  Let  the  polygon  described  about  the  circle  be  called  M,  and  the 
polygon  inscribed  be  called  N ;  then,  because  these  polygons  are  similar, 


168  SUPPLEMENT  TO  THE  ELEMENTS 


they  are  as  tlie  squares  of  the  homologous  sides  HL  and  AF  (3.  Corol. 
20.  6.),  that  is,  because  the  triangles  HLG,  AFG  are  similar,  as  the  square 
of  HG  to  the  square  of  AG,  that  is  of  GK.  But  the  triangles  HGK,  ACF 
have  been  proved  to  be  similar,  and  therefore  the  square  of  AC  is  to  the 
square  of  CF  as  the  polygon  M  to  the  polygon  N  ;  and,  by  conversion, 
the  square  of  AC  is  to  its  excess  above  the  squares  of  CF,  that  is,  to  the 
square  of  AF  (47.  1.),  as  the  polygon  M  to  its  excess  above  the  polygon 
N.  But  the  square  of  AC,  that  is,  the  square  described  about  the  circle 
ABC  is  greater  than  the  equilateral  polygon  of  eight  sides  described  about 
the  circle,  because  it  contains  that  polygon  ;  and,  for  the  same  reason,  the 
polygon  of  eight  sides  is  greater  than  the  polygon  of  sixteen,  and  so  on ; 
therefore,  the  square  of  AC  is  greater  than  any  pojygon  described  about 
the  circle  by  the  continual  bisection  of  the  arc  AB  ;  it  is  therefore  greater 
than  the  polygon  M.  Now,  it  has  V.en  demonstrated,  that  the  square  of 
AC  is  to  the  square  of  AF  as  the  polygon  M  to  the  difference  of  the  poly- 
gons ;  therefore,  since  the  square  of  AC  is  greater  than  M,  the  square  of 
AF  is  greater  than  the  difference  of  the  polygons  (14.  5.).  The  difference 
of  the  polygons  is  therefore  less  than  the  square  of  AF  ;  but  AF  is  less 
than  D  ;  therefore  the  difference  of  the  polygons  is  less  than  the  square  of 
D  ;  that  is,  than  the  given  space. 

CoR.  1.  Because  the  polygons  M  and  N  differ  from  one  another  more 
than  either  of  them  differs  from  the  circle,  the  difference  between  each  of 
them  and  the  circle  is  less  than  the  given  space,  viz.  the  square  of  D.  And 
therefore,  however  small  any  given  space  may  be,  a  polygon  may  be  in- 
scribed in  the  circle,  and  another  described  about  it,  each  of  which  shall 
differ  from  the  circle  by  a  space  less  than  the  given  space. 

CoR.  2.  The  space  B,  which  is  greater  than  any  polygon  that  can  be 
inscribed  in  the  circle  A,  and  less  than  any  polygon  that  can  be  described 
about  it,  is  equal  to  the  circle  A.  If  not,  let  them  be  unequal ;  and  first, 
let  B  exceed  A  by  the  space  C.  Then,  because  the  polygons  described 
about  the  circle  A  are  all  greater  than  D,  by  hypothesis  ;  and  because  B 
is  greater  than  A  by  the  space  C,  therefore  no  polygon  can  be  described 


OF  GEOMETRY.    BOOK  I. 


169 


about  the  circle  A,  but  what  must  exceed  it  by  a  space  greater  than  C, 
which  is  absurd.  In  the  same  manner,  if  B  be  less  than  A  by  the  space 
C,  it  is  shewn  that  no  polygon  can  be  inscribed  in  the  circle  A,  but  what 
is  less  than  A  by  a  space  greater  than  C,  which  is  also  absurd.  Therefor^, 
A  and  B  are  not  unequal ;  that  is,  they  are  equal  to  one  another. 

PROP.  V.    THEOR. 

The  area  of  any  circle  is  equal  to  the  rectangle  contained  hy  the  semi-diameter, 
and  a  straight  line  equal  to  half  the  circumference. 

Let  ABC  be  a  circle  of  which  the  centre  is  D,  and  the  diameter  AC  ;  if 
in  AC  produced  there  be  taken  AH  equal  to  half  the  circumference,  the 
area  of  the  circle  is  equal  to  the  rectangle  contained  by  DA  and  AH. 

Let  AB  be  the  side  of  any  equilateral  polygon  inscribed  in  the  circle 
ABC  ;  bisect  the  circumference  AB  in  G,  and  through  G  draw  EGF 
touching  the  circle,  and  meeting  DA  produced  in  E,  and  DB  produced  in 


Kn  L 


F  ;  EF  will  be  the  side  of  an  equilateral  polygon  described  about  the  cir- 
cle ABC  (3.  L  Sup.).  In  AC  produced  take  AK  equal  to  half  the  peri- 
meter of  the  polygon  whose  side  is  AB  ;  and  AL  equal  to  half  the  perime- 
ter of  the  polygon,  whose  side  is  EF.  Then  AK  will  be  less,  and  AL 
greater  than  the  straight  line  AH  (Lera.  Sup.).  Now,  because  in  the 
triangle  EDF,  DG  is  drawn  perpendicular  to  the  base,  the  triangle  EDF 

22 


170  SUPPLEMENT  TO  THE  ELEMENTS 

is  equal  to  the  rectangle  contained  by  DG  and  the  half  of  EF  (41.  I.) ;  and 
as  the  same  is  true  of  all  the  other  equal  triangles  having  their  vertices  in 
D,  which  make  up  the  polygon  described  about  the  circle  ;  therefore,  the 
whole  polygon  is  equal  to  the  rectangle  contained  by  DG  and  AL,  half  the 
perimeter  of  the  polygon  (L  2.),  or  by  DA  and  AL.  But  AL  is 
greater  than  AH,  therefore  the  rectangle  DA.AL  is  greater  than  the  rect- 
angle DA.AH  ;  the  rectangle  DA.AH  is  therefore  less  than  the  rectangle 
DA.AL,  that  is,  than  any  polygon  described  about  the  circle  ABC. 

Again,  the  triangle  ADB  is  equal  to  the  rectangle  contained  by  DM  the 
perpendicular,  and  one  half  of  the  base  AB,  and  it  is  therefore  less  than  the 
rectangle  contained  by  DG,  or  DA,  and  the  half  of  AB      And  as  the  same 


is  true  of  all  the  other  triangles  having  their  vertices  in  D,  which  make 
up  the  inscribed  polygon,  therefore  the  whole  of  the  inscribed  polygon  is 
less  than  the  rectangle  contained  by  DA,  and  AK  half  the  perimeter  of  the 
polygon.  Now,  the  rectangle  DA.AK  is  less  than  DA.AH  ;  much  more, 
therefore,  is  the  polygon  whose  side  is  AB  less  than  DA.AH ;  and  the 
rectangle  Dx\.AH  is  therefore  greater  than  any  polygon  inscribed  in  the 
circle  ABC.  But  the  same  rectangle  DA.AH  has  been  proved  to  be  less 
than  any  polygon  described  about  the  circle  ABC  ;  therefore  the  rectangle 
DA.AH  is  equal  to  the  circle  ABC  (2.  Cor.  4.  1.  Sup.).  Now  DA  is  the 
semidiameter  of  the  circle  ABC,  and  AH  the  half  of  its  circumference. 

CoR.  1.  Because  DA  :  AH  : :  DA^  :  DA.AH  (1.  6.),  and  because  by 
this  proposition,  DA.AH  =  the  area  of  the  circle,  of  which  DA  is  the  ra- 
dius :  therefore,  as  the  radius  of  any  circle  to  the  semicircumference,  or  as 
the  diameter  to  the  whole  circumference,  so  is  the  square  of  the  radius  to 
the  area  of  the  circle. 

Cor.  .2.  Hence  a  polygon  may  be  described  about  a  circle,  the  perime- 
ter of  which  shall  exceed  the  circumference  of  the  circle  by  a  line  that  is 
less  than  any  given  line.  Let  NO  be  the  given  line.  Take  in  NO  the 
part  NP  less  than  its  half,  and  also  than  AD,  and  let  a  polygon  be  describ- 
ed about  the  circle  ABC,  so  that  its  excess  above  ABC  may  be  less  than 
the  square  of  NP  (1.  Cor.  4.  1.  Sup.).  Let  the  side  of  this  polygon  be  EF. 
And  since,  as  has  been  proved,  the  circle  is  equal  to  the  rectangle  DA.AH, 
and  the  polygon  to  the  rectangle  DA.AL,  the  excess  of  the  polygon  above 
the  circle  is  equal  to  the  rectangle  DA.HL  j  therefore  the  rectangle  DA. 


OF  GEOMETRY.    BOOK  I. 


171 


HL  is  less  than  the  square  of  NP  ;  and  therefore,  since  DA  is  greater  than 
ICP,  HL  is  less  than  NP,  and  twice  HL  less  than  twice  NP,  wherefore, 
much  more  is  twice  HL  less  than  NO.  But  HL  is  the  difference  between 
half  the  perimeter  of  the  polygon  whose  side  is  EF,  and  half  the  circum- 
ference of  the  circle  ;  therefore,  twice  HL  is  the  difference  between  the 
whole  perimeter  of  the  polygon  and  the  whole  circumference  of  the  circle 
(5.  5.).  The  difference,  therefore,  between  the  perimeter  of  the  polygon 
and  the  circumference  of  the  circle  is  less  than  the  given  line  NO. 

Cor.  3.  Hence,  also,  a  polygon  may  be  inscribed  in  a  circle,  such 
that  the  excess  of  the  circumference  above  the  perimeter  of  the  polygon 
may  be  less  than  any  given  line.     This  is  proved  like  the  preceding. 

PROP.  VL    THEOR. 

The  areas  of  circles  are  to  one  another  in  the  duplicate  ratio,  or  as  the  squares 

of  their  diameters. 

Let  ABD  and  GHL  be  two  circles,  of  which  the  diameters  are  AD  and 
GL ;  the  circle  ABD  is  to  the  circle  GHL  as  the  square  of  AD  to  the 
square  of  GL. 

Let  ABCDEF  and  GHKLMN  be  two  equilateral  polygons  of  the  same 
number  of  sides  inscribed  in  the  circles  ABD,  GHL  ;  and  let  Q  be  such  a 


DG 


space  that  the  square  of  AD  is  to  the  square  of  GL  as  the  circle  ABD  to 
the  space  Q.  Because  the  polygons  ABCDEF  and  GHKLMN  are  equi- 
lateral and  of  the  same  number  of  sides,  they  are  similar  (2.  1.  Sup.),  and 


172  SUPPLEMENT  TO  THE  ELEMENTS 

tlieir  areas  are  as  the  squares  of  the  diameters  of  the  circles  in  which  they 
are  inscribed.  Therefore  AD^  :  GL^  : :  polygon  ABCDEF  ;  polygon 
GHKLMN;  butAD2  :  GL2  : :  circle  ABD  :  Q  ;  and  therefore,  ABCDEF 
.  GHKLM  : :  circle  ABD  :  Q.  Now,  circle  ABD /ABCDEF;  there- 
fore Q/ GHKLMN  (14.  5.),  that  is,  Q  is  greater  than  any  polygon  in- 
scribed in  the  circle  GHL. 

In  the  same  manner  it  is  demonstrated,  that  Q  is  less  than  any  polygon 
described  about  the  circle  GHL  ;  wherefore  the  space  Q  is  equal  to  the 
circle  GHL  (2.  Cor.  4.  1.  Sup.).  Now,  by  hypothesis,  the  circle  ABD  is 
to  the  space  Q  as  the  square  of  AD  to  the  square  of  GL  ;  therefore  the 
circle  ABD  is  to  the  circle  GHL  as  the  square  of  AD  to  the  square  of  GL. 

CoR.  1.  Hence  the  circumferences  of  circles  are  to  one  another  as 
their  diameters. 

Let  the  straight  line  X  be  equal  to  half  the  circumference  of  the  circle 
ABD,  and  the  straight  line  Y  to  half  the  circumference  of  the  circle  GHL: 

X ■■ 


And  because  the  rectangles  AO.X  and  GP.Y  are  equal  to  the  circles  ABD 
and  GHL  (5.  1.  Sup.),  therefore  AO.X  :  GP.Y  : :  AD^  :  GL^  : :  AO2  : 
GP2 ;  and  alternately,  AO.X  :  AO^  :  :  GP.Y  :  GP^ ;  whence,  because 
rectangles  that  have  equal  altitudes  are  as  their  bases  (1.  6.),  X  :  AO  :  : 
Y  ;  GP,  and  again  alternately,  X  :  Y  :  :  AO  :  GP :  wherefore,  taking  the 
doubles  of  each,  the  circumference  ABD  is  to  the  circumference  GHL  as 
the  diameter  AD  to  the  diameter  GL. 

CoR.  2.  The  circle  that  is  described  upon  the  side  of  a  right  angled 
triangle  opposite  to  the  right  angle,  is  equal  to  the  two  circles  described  on 
the  other  two  sides.  For  the  circle  described  upon  SR  is  to  the  circle  de- 
scribed upon  RT  as  the  square  of  SR  to  the  square  of  RT  ;  and  the  circle 
described  upon  TS  is  to  the  circle  described  upon  RT  as  the  square  of  ST 
to  the  square  of  RT.  Wherefore, 
the  circles  described  on  SR  and  on 
ST  are  to  the  circle  described  on  RT 
as  the  squares  of  SR  and  of  ST  to 
the  square  of  RT  (24.  5.).  But  the 
squares  of  RS  and  of  ST  are  equal 
to  the  square  of  RT  (47.  1.) ;  there- 
fore the  circles  described  on  RS  and 
ST  are  equal  to  the  circle  described 
on  RT. 

PROP.  VIL     THEOR. 

Equiangular  parallelograms  are  to  one  another  as  the  products  of  the  num 
bers  proportional  to  their  sides. 

Let  AC  and  DF  be  two  equiangular  parallelograms,  and  let  M,  N,  P 
and  Q  be  four  numbers,  such  that  AB  :  BC  : :  M  :  N ;  AB  :  DE  : :  M  : 


OF  GEOMETRY.     BOOK  I.  173 

P ;  and  AB  :  EF  : :  M  :  Q,  and  therefore  ex  ^quali,  BC  :  EF  : :  N  :  Q. 
The  parallelogram  AC  is  to  the  parallelogram  DF  as  MN  to  PQ. 

Let  NP  be  the  product  of  N  into  P,  and  the  ratio  of  MN  to  PQ  will  be 
compounded  of  the  ratios  (def.  10.  5.)  of  MN  to  NP,  and  NP  to  PQ. 
But  the  ratio  of  MN  to  NP  is  the  same  with  that  of  M  to  P  (15.  5.),  be- 


iC 


A.  B     D  E 

cause  MN  and  NP  are  equimultiples  of  M  and  P  ;  and  for  the  same  reason, 
the  ratio  of  NP  to  PQ  is  the  same  with  that  of  N  to  Q  ;  therefore  the  ratio 
of  MN  to  PQ  is  compounded  of  the  ratios  of  M  to  P,  and  of  N  to  Q.  Now, 
the  ratio  of  M  to  P  is  the  same  with  that  of  the  side  AB  to  the  side  DE  (by- 
Hyp.)  ;  and  the  ratio  of  N  to  Q  the  same  with  that  of  the  side  BC  to  the 
side  EF.  Therefore,  the  ratio  of  MN  to  PQ  is  compounded  of  the  ratios 
of  AB  to  DE,  and  of  BC  to  EF.  And  the  ratio  of  the  parallelogram  AC 
to  the  parallelogram  DF  is  compounded  of  the  same  ratios  (23.  6.) ;  there- 
fore, the  parallelogi-am  AC  is  to  the  parallelogram  DF  as  MN,  the  product 
of  the  numbers  M  and  N,  to  PQ,  the  product  of  the  numbers  P  and  Q. 

Cor.  1.  Hence,  if  GH  be  to  KL  as  the  number  M  to  the  number  N  ; 
the  square  described  on  GH  will  be  to 

the  square  described  on  KL  as  MM,  the     Q  jj  j^  L 

square  of  the   number   M  to   NN,  the 
square  of  the  number  N. 

Cor.  2.  If  A,  B,  C,  D,  &c.  are  any  lines,  and  m,  n,  r,  5,  &c.  numbers 
proportional  to  them ;  viz.  A  :  B  :  :  wi  :  n,  A  :  C  : :  ot  :  r,  A  :  D  :  :  m :  «•, 
&c. ;  and  if  the  rectangle  contained  by  any  two  of  the  lines  be  equal  to  the 
square  of  a  third  line,  the  product  of  the  numbers  proportional  to  the  first 
two,  will  be  equal  to  the  square  of  the  number  proportional  to  the  third , 
that  is,  if  A.C=B2,  »iXr=nXw,  or=w2. 

For  by  this  Prop.  A.C  :  B^  :  :  mXr  :  n^  ;  but  A.C=B2,  therefore  mXr 
=n2.  Nearly  in  the  same  way  it  may  be  demonstrated,  that  whatever  is 
the  relation  between  the  rectangles  contained  by  these  lines,  there  is  the 
same  between  the  products  of  the  numbers  proportional  to  them. 

So  also  conversely  if  m  and  r  be  numbers  proportional  to  the  lines  A  and 
C  ;  if  also  A.C=B2,  and  if  a  number  n  be  found  such,  that  n^=mry  then 
A  :  B  :  :  771 :  n.  For  let  A  :  B  : :  m  :  q,  then  since  m,  q,  r  are  proportional 
to  A,  B,  and  C,  and  A.C=B2 ;  therefore,  as  has  just  been  proved,  q^=m 
Xr;  but  n'^=:qxr,  by  hypothesis,  therefore  n'^=q'^,  and  n=q ;  wherefore 
A  :  B  :  :  m  :  n. 

SCHOLIUM. 
In  order  to  have  numbers  proportional  to  any  set  of  magnitudes  of  the 


174 


SUPPLEMENT  TO  THE  ELEMENTS 


same  kind,  suppose  one  of  them  to  be  divided  into  any  number  m,  of  equal 
parts,  and  let  H  be  one  of  those  parts.  Let  H  be  found  n  times  in  the  mag- 
nitude B,  r  times  in  C,  ^  times  in  D,  &c.,  then  it  is  evident  that  the  num- 
bers m,  Hy  r,  s  are  proportional  to  the  magnitudes  A,  B,  C  and  D.  When 
therefore  it  is  said  in  any  of  the  following  propositions,  that  a  line  as  A= 
a  number  m,  it  is  imderstood  that  A=77iX  H,  or  that  A  is  equal  to  the  given 
magnitude  H  multiplied  by  m,  and  the  same  is  understood  of  the  other 
magnitudes,  B,  C,  D,  and  their  proportional  numbers,  H  being  the  common 
measure  of  all  the  magnitudes.  This  common  measure  is  omitted  for  the 
sake  of  brevity  in  the  arithmetical  expression ;  but  is  always  implied,  when 
a  line,  or  other  geometrical  magnitude,  is  said  to  be  equal  to  a  number. 
Also,  when  there  are  fractions  in  the  number  to  which  the  magnitude  is 
called  equal,  it  is  meant  that  the  common  measure  H  is  farther  subdivided 
into  such  parts  as  the  numerical  fraction  indicates.  Thus,  if  A=360.375, 
it  is  meant  that  there  is  a  certain  magnitude  H,  such  that  A=360xH4- 

375 

X  H,  or  that  A  is  equal  to  360  times  H,  together  with  375  of  the 

thousandth  parts  of  H.  And  the  same  is  true  in  all  other  cases,  where 
numbers  are  used  to  express  the  relations  of  geometrical  magnitudes. 


PROP.  Vin.    THEOR. 

The  perpendicular  drawn  from  the  centre  of  a  circle  on  the  chord  of  any  arc  tsa 
mean  proportional  between  half  the  radius  and  the  line  made  up  of  the  radius 
and  the  perpendicular  drawn  from  the  centre  on  the  chord  of  double  that  arc : 
And  the  chord  of  the  arc  is  a  mean  proportional  between  the  diameter  and  a  line 
which  is  the  difference  between  the  radius  and  the  aforesaid  perpendicular  from 
the  centre 

Let  ADB  be  a  circle,  of  which  the  centre  is  C  ;  DBE  any  arc,  and  DB 
the  half  of  it ;  let  the  chords  DE,  DB  be  drawn  :  as  also  CF  and  CG  at 
right  angles  to  DE  and  DB  ;  if  CF  be  produced  it  will  meet  the  circum 
ference  in  B ;  let  it  meet  it  again  in  A,  and  let  AC  be  bisected  in  H ;  CG 


OF  GEOMETRY.    BOOK  I. 


175 


is  a  mean  proportional  between  AH  and  AF  ;  and  BD  a  mean  proportional 
between  AB  and  BF,  the  excess  of  the  radius  above  CF. 

Join  AD  ;  and  because  ADB  is  a  right  angle,  being  an  angle  in  a  semi- 
circle ;  and  because  CGB  is  also  a  right  angle,  the  triangles  ABD,  CBG 
are  equiangular,  and,  AB  :  AD  : :  BC  :  CG  (4.  6.),  or  alternately,  AB  : 
BC  : :  AD  :  CG  ;  and  therefore,  because  AB  is  double  of  BC,  AD  is  dou- 
ble of  CG,  and  the  square  of  AD  therefore  equal  to  four  times  the  square 
ofCG. 

But,  because  ADB  is  a  right  angled  triangle,  and  DF  a  perpendicular 
on  AB,  AD  is  a  mean  proportional  between  AB  and  AF  (8.  6.),  and  AD^ 
=AB.AF  (17.  6.),  or  since  AB  is  =4AH,  AD2=4AH.AF.  Therefore 
also,  because  4CG2=:AD2,  4CG2=4AH.AF,  and  CG2=AH.AF  ;  where 
fore  CG  is  a  mean  proportional  between  AH  and  AF,  that  is,  between  half 
the  radius  and  the  line  made  up  of  the  radius,  and  the  perpendicular  on  the 
chord  of  twice  the  arc  BD. 

Again,  it  is  evident  that  BD  is  a  mean  proportional  between  AB  and  3F 
(8.  6.),  that  is,  between  the  diameter  and  the  excess  of  the  radius  above 
the  perpendicular,  on  the  chord  of  twice  the  arc  DB. 


PROP.  IX.    THEOR.* 

The  circumference  of  a  circle  exceeds  three  times  the  diameter y  hy  a  line  less 
than  ten  of  the  parts,  of  which  the  diameter  contains  seventy,  but  greater 
than  ten  of  the  parts  whereof  the  diameter  contains  seventy-one. 

Let  ABD  be  a  circle,  of  which  the  centre  is  C,  and  the  diameter  AB  ; 
the  circumference  is  greater  than  three  times  AB,  by  a  line  less  than  — ,  or 


70' 


-,  of  AB,  but  greater  than  — -  of  AB. 


*  In  this  proposition,  the  character  -|-  placed  after  a  number,  signliles  that  something  is  to 
be  added  to  it ;  and  the  character  — ,  on  the  other  hand,  signifies  that  something  is  to  be  taken 
away  from  it. 


176  SUPPLEMENT  TO  THE  ELEMENTS 

In  the  circle  ABD  apply  the  straight  line  BD  equal  to  the  radius  BC : 
Draw  DF  perpendicular  to  BC,  and  let  it  meet  the  circumference  again  in 
E  ;  draw  also  CG  perpendicular  to  BD  :  produce  BC  to  A,  bisect  AC  in 
H,  and  join  CD. 

It  is  evident,  that  the  arcs  BD,  BE  are  each  of  them  one-sixth  of  the 
circumference  (Cor.  15.  4.),  and  that  therefore  the  arc  DBE  is  one  third  of 
the  circumference.  Wherefore,  the  line  (8.  1.  Sup.)  CG  is  a  mean  pro- 
portional between  AH,  half  the  radius,  and  the  line  AF.  Now  because  the 
sides  BD,  DC,  of  the  triangle  BDC  are  equal,  the  angles  DCF,  DBF  are 
also  equal ;  and  the  angles  DFC,  DFB  being  equal,  and  the  side  DF  com- 
mon to  the  triangles  DBF,  DCF,  the  base  BF  is  equal  to  the  base  CF,  and 
BC  is  bisected  in  F. 

Therefore,  if  AC  or  BC=1000,  AH=:500,  CF=500,  AF=1500,  and 
CG  being  a  mean  proportional  between  AH  and  AF,  CG2=(17.  6.)  AH. 
AF=500x  1500=750000;  wherefore  CG=866.0254+,  because  (866. 
0254)2  is  less  than  750000.     Hence  also,  AC+CG=1866.0254-f . 

Now,  as  CG  is  the  perpendicular  drawn  from  the  centre  C,  on  the  chord 
of  one-sixth  of  the  circumference,  if  P  =  the  perpendicular  from  C  on  the 
chord  of  one-twelfth  of  the  circumference,  P  will  be  a  mean  proportional 
between  AH  (8.  1.  Sup.)  and  AC+CG,  and  P2=AH  (AC-fCG)= 
500  X  (1866.0254-f)  =  933012.7+.  Therefore,  P  =  965.9258-f,  be- 
cause (965.9258)2  is  less  than  933012.7.  Hence  also,  AC-fP=1965. 
9258-f-. 

Again,  if  Q  =  the  perpendicular  drawn  from  C  on  the  chord  of  one 
twenty-fourth  of  the  circumference,  Q  will  be  a  mean  proportional  between 
AH  and  AC+P,  and  Q2=AH  (AC-f  P)=500(1965.9258+)=982962. 
9+  ;  and  therefore  Q=991.4449-f ,  because  (991.4449)2  is  less  than 
982962.9.     Therefore  also  AC-f  Q=1991.4449-f . 

In  like  manner,  if  S  be  the  perpendicular  from  C  on  the  chord  of  one 
forty-eighth  of  the  circumference,  S2=AH  (AC-i-Q)=500  (1991.4449+) 
=995722.45+;  and  S=997.8589+,  because  (997.8589)2  is  less  than 
995722.45.     Hence  also,  AC+S=1997.8589+. 

Lastly,  if  T  be  the  perpendicular  from  C  on  the  chord  of  one  ninety-sixth 
of  the  circumference,  T2=AH  (AC+S)=500  (1997.8589+)=998929. 
45+,  and  T=999.46458+.  Thus  T,  the  perpendicular  on  the  chord  of 
one  ninety-sixth  of  the  circumference,  is  greater  than  999.46458  of  those 
parts  of  which  the  radius  contains  1000. 

But  by  the  last  proposition,  the  chord  of  one  ninety-sixth  part  of  the  cir- 
cumference is  a  mean  proportional  between  the  diameter  and  the  excess  of 
the  radius  above  S,  the  perpendicular  from  the  centre  on  the  chord  of  one 
forty-eighth  of  the  circumference.  Therefore,  the  square  of  the  chord  of 
one  ninety-sixth  of  the  circumference =AB  (AC— S)  =2000  X  (2.1411—,) 
=4282.2—;  and  therefore  the  chord  itself  =65.4386— ,  because  (65. 
4386)2  is  greater  than  4282.2.  Now,  the  chord  of  one  ninety-sixth  of  the 
circumference,  or  the  side  of  an  equilateral  polygon  of  ninety-six  sides  in- 
scribed in  the  circle,  being  65.4386—,  the  perimeter  of  that  polygon  will  be 
=(65.4386—)  96=6282.1056—. 

Let  the  perimeter  of  the  circumscribed  polygon  of  the  same  number  of 
sides,  be  M,then  (2.  Cor.  2.  l.Sup.)T  :  AC  : :  6282.1056—  :  M,  that  is, 
(since  T=999.46458  +  ,  as  already  shewn), 


I 


i 


OF  GEOMETRY.    BOOK  I. 


177 


999.46458+  :  1000  :  :  6282.1056—  :  M ;  if  then  N  be  such, 
that 999.46458  :  1000  :  :  6282.1056—  :  N;  ex  sequo  perturb.  999.46458 
4-  :  999.46458  : :  N  :  M  ;  and,  since  the  first  is  greater  than  the  second, 
the  third  is  greater  than  the  fourth,  or  N  is  greater  than  M. 

Now,  if  a  fourth  proportional  be  found  to  999.46458,  1000  and  6282. 
1056  viz.  6285.461—,  then, 

because,  999.46458  :  1000  : :  6282.1056  :  6285.461—, 
and  as  before,  999.46458  :  1000  : :  6282.1056—  :  N  ; 
therefore,  6282.1056  :  6282.1056—  :  :  6285.461— N,  and  as  the  first  of 
these  proportionals  is  greater  than  the  second,  the  third,  ^viz.  6285  461  — 


is  greater  than  N,  the  fourth.  But  N  was  proved  to  be  greater  than  M  ; 
much  more,  therefore,  is  6285.461  greater  than  M,  the  perimeter  of  a  poly- 
gon of  ninety-six  sides  circumscribed  about  the  circle  ;  that  is,  the  perime- 
ter of  that  polygon  is  less  than  6285.461  ;  now,  the  circumference  of  the 
circle  is  less  than  the  perimeter  of  the  polygon  ;  much  more,  therefore,  is  it 
less  than  6285.461  ;  wherefore  the  circumference  of  a  circle  is  less  than 
6285.461  of  those  parts  of  which  the  radius  contains  1000.  The  circum- 
ference, therefore  has  to  the  diameter  a  less  ratio  (8.  5.)  than  6285.461  has 
to  2000,  or  than  3142.7305  has  to  1000  :  but  the  ratio  of  22  to  7  is  greater 
than  the  ratio  of  3142.7305  to  1000,  therefore  the  circumference  has  a  less 
ratio  to  the  diameter  than  22  has  to  7,  or  the  circumference  is  less  than  22 
of  the  parts  of  which  the  diameter  contains  7. 

It  remains  to  demonstrate,  that  the  part  by  which  the  circumference  ex- 
ceeds the  diameter  is  greater  than  ~-  of  the  diameter. 

It  was  before  shewn,  that  00^=750000;  wherefore  00=866.02545-, 
because  (866.02545)2  is  greater  than  750000  ;  therefore  AO-f  00=1866. 
02545—. 

Now,  P  being,  as  before,  the  perpendicular  from  the  centre  on  the  chord 
of  one  twelfth  of  the  circumference,  P2=AH  (AO+CG)  =-500  X  (1866. 
02545)— =933012.73— ;  and  P  =  965.92585— ,  because  (965.92585)2 
is  greater  than  633012.73.     Hence  also,  AC +  P= 1965.92585—. 

23 


178  SUPPLEMENT  TO  THE  ELEMENTS 

Next,  as  Q=  the  perpendicular  drawn  from  the  centre  on  the  chord  of 
one  twenty-fourth  of  the  circumference,  Q2=AH(AC-|-P)=o00x  (1965. 
92585—)  =982962.93— i  and  Q  =  991.44495—,  because  (991.44496)2 
is  greater  than  982962.93.     Hence  also,  AC+Q=1991.44495— . 

In  like  manner,  as  S  is  the  perpendicular  from  C  on  the  chord  of  one 
forty-eighth  of  the  circumference,  S2=AH  (AC+Q)=500(1991.44495— ) 
=995722.475—,  and  S=(997.85895—)  because  (997.85895)2  is  greater 
than  995722.475. 

But  the  square  of  the  chord  of  the  ninety-sixth  part  of  the  circumference 
=AB  (AC— S)=2000  (2.14105-f)=4282.14-,  and  the  chord  itself  = 
65.4377 -f  because  (65.4377)2  is  less  than  4282.1  :  Now  the  chord  of  one 
ninety-sixth  part  of  the  circumference  being  =65.4377+,  the  perimeter 
of  a  polygon  of  ninety-six  sides  inscribed  in  the  circle  =(65.4377-|-)96= 
6282.01 9-j-.  But  the  circumference  of  the  circle  is  greater  than  the  pe- 
rimeter of  the  inscribed  polygon ;  therefore  the  circumference  is  greater 
than  6282.019,  of  those  parts  of  which  the  radius  contains  1000  ;  or  than 
3141.009  of  the  parts  of  which  the  radius  contains  500,  or  the  diameter 

contains  1000.     Now,  3141.009  has  to  1000  a  greater  ratio  than  3-f  ~ 

to  1  ;  therefore  the  circumference  of  the  circle  has  a  greater  ratio  to  the 

diameter  than  34-  =t-has  to  1  :  that  is,  the  excess  of  the  circumference 
71 

above  three  times  the  diameter  is  greater  than  ten  of  those  parts  of  which 

the  diameter  contains  71  ;  and  it  has  already  been  shewn  to  be  less  than 

ten  of  those  of  which  the  diameter  contains  70. 

CoR.  1.  Hence  the  diameter  of  a  circle  being  given,  the  circumference 
may  be  found  nearly,  by  making  as  7  to  22,  so  the  given  diameter  to  a 
fourth  proportional,  which  will  be  greater  than  the  circumference.     And 

if  as  1  to  3  4-  7^1  or  as  71  or  223,  so  the  given  diameter  to  a  fourth  pro- 
portional, this  will  be  nearly  equal  to  the  circumference,  but  will  be  less 
than  it. 

Cor.  2.     Because  the  difference  between  -  and  — -  is  — -,  therefore  the 

lines  found  by  these  proportionals  differ  by  -—  of  the  diameter.  There- 
fore the  difference  of  either  of  them  from  the  circumference  must  be  less 
than  the  497th  part  of  the  diameter. 

CoR.  3.  As  7  to  22,  so  the  square  of  the  radius  to  the  area  of  the  circle 
nearly. 

For  it  has  been  shewn,  that  (1.  Cor.  5.  1.  Sup.)  the  diameter  of  a  cir- 
cle is  to  its  circumference  as  the  square  of  the  radius  to  the  area  of  the 
circle  ;  but  the  diameter  is  to  the  circumference  nearly  as  7  to  22,  there- 
fore the  square  of  the  radius  is  to  the  area  of  the  circle  nearly  in  that  same 
ratio. 


OF  GEOMETRY.    BOOK  I. 


179 


SCHOLIUM. 

It  is  evident  that  the  method  employed  in  this  proposition,  for  finding 
the  limits  of  the  ratio  of  the  circumference  of  the  diameter,  may  be  carried 
to  a  greater  degree  of  exactness,  by  finding  the  perimeter  of  an  inscribed 
and  of  a  circumscribed  polygon  of  a  greater  number  of  sides  than  96.  The 
manner  in  which  the  perimeters  of  such  polygons  approach  nearer  to  one 
another,  as  the  number  of  their  sides  increases,  may  be  seen  from  the  fol- 
lowing Table,  which  is  constructed  on  the  principles  explained  in  the  fore- 
going Proposition,  and  in  which  the  radius  is  supposed  =1. 


NO.  of  Sides 

Perimeter  of  the 

Perimeter  of  the 

of  the  Poly- 

inscribed Poly- 

circumscribed 

gon. 

gon. 

Polygon. 

6 

6.000000 

6.822033— 

12 

6.211657-f 

6.430781  — 

24 

6.265257+ 

6.319320— 

48 

6.278700+ 

6.292173— 

96 

6.282063  + 

6.285430— 

192 

6,282904+ 

6.283747- 

384 

6.283115+ 

6.283327— 

768 

6.283167+ 

6.283221  — 

1536 

6.283180+ 

6.283195- 

3072 

6.283184+ 

6.283188— 

6144 

6.283185+ 

6.283186— 

The  part  that  is  wanting  in  the  numbers  of  the  second  column,  to  make 
up  the  entire  perimeter  of  any  of  the  inscribed  polygons,  is  less  than  unit 
in  the  sixth  decimal  place ;  and  in  like  manner,  the  part  by  which  the 
numbers  in  the  last  column  exceed  the  perimeter  of  any  of  the  circumscrib- 
ed polygons  is  less  than  a  unit  in  the  sixth  decimal  place,  that  is,  than 

1  QOQono  °^  ^^®  radius.     Also,  as  the  numbers  in  the  second  column  are 

less  than  the  perimeters  of  the  inscribed  polygons,  they  are  each  of  them 
less  than  the  circumference  of  the  circle  ;  and  for  the  same  reason,  each  of 
those  in  the  third  column  is  greater  than  the  circumference.     But  when 

the  arc  of  -  of  the  circumference  is  bisected  ten  times,  the  number  of  sides 
in  the  polygon  is  6144,  and  the  numbers  in  the  Table  diff*er  from  one  an- 
other only  by  part  of  the  radius,  and  therefore  the  perimeters  of 

the  polygons  difier  by  less  than  that  quantity ;  and  consequently  the  cir- 
cumference of  the  circle,  which  is  greater  than  the  least,  and  less  than  the 
greatest  of  these  numbers,  is  determined  within  less  than  the  millionth 
part  of  the  radius. 

Hence  also,  if  R  be  the  radius  of  any  circle,  the  circumference  is  greater 
than  Rx  6.283185,  or  than  2Rx  3.141592,  but  less  than  2Rx  3.141593  ; 


180  SUPPLEMENT  TO  THE  ELEMENTS,  &c. 

and  these  numbers  differ  from  one  another  only  by  a  millionth  part  of  the 
radius.  So  also  R2+3.141592  is  less,  and  R2  x  3.141593  greater  than  the 
area  of  the  circle  ;  and  these  numbers  differ  from  one  another  only  by  a 
millionth  part  of  the  square  of  the  radius. 

In  this  way,  also,  the  circumference  and  the  area  of  the  circle  may  be 
found  still  nearer  to  the  truth ;  but  neither  by  this,  nor  by  any  other  me- 
thod yet  known  to  geometers,  can  they  be  exactly  determined,  though  the 
errors  of  both  may  be  reduced  to  a  less  quantity  than  any  that  can  be  as- 
signed. 


ELEMENTS 

OF 

GEOMETRY. 


SUPPLEMENT. 


BOOK  11. 

OF  THE  INTERSECTION  OF  PLANES. 


DEFINITIONS. 


1.  A  STRAIGHT  line  is  perpendicular  or  at  right  angles  to  a  plane,  when 
it  makes  right  angles  with  every  straight  line  which  it  meets  in  that 
plane. 

2.  A  plane  is  perpendicular  to  a  plane,  when  the  straight  lines  drawn  in 
one  of  the  planes  perpendicular  to  the  common  section  of  the  two  planes, 
are  perpendicular  to  the  other  plane. 

3.  The  inclination  of  a  straight  line  to  a  plane  is  the  acute  angle  contained 
by  that  straight  line,  and  another  drawn  from  the  point  in  which  the 
first  line  meets  the  plane,  to  the  point  in  which  a  perpendicular  to  the 
plane,  drawn  from  any  point  of  the  first  line,  meets  the  same  plane. 

4.  The  angle  made  by  two  planes  which  cut  one  another,  is  the  angle  con- 
tained by  two  straight  lines  drawn  from  any,  the  same  point  in  the  line 
of  their  common  section,  at  right  angles  to  that  line,  the  one,  in  the  one 
plane,  and  the  other,  in  the  other.  Of  the  two  adjacent  angles  made  by 
two  lines  drawn  in  this  manner,  that  which  is  acute  is  also  called  the  in- 
clination of  the  planes  to  one  another. 

5.  Two  planes  are  said  to  have  the  same,  or  a  like  inclination  to  one  an- 
other, which  two  other  planes  have,  when  the  angles  of  inclination  above 
defined  are  equal  to  one  another. 

6.  A  straight  line  is  said  to  be  parallel  to  a  plane,  when  it  does  not  meet 
the  plane,  though  producred  ever  so  far. 


182  SUPPLEMENT  TO  THE  ELEMENTS 

7.  Planes  are  said  to  be  parallel  to  one  another,  which  do  not  meet,  though 
produced  ever  so  far. 

8.  A  solid  angle  is  an  angle  made  by  the  meeting  of  more  than  two  plane 
angles,  which  are  not  in  the  same  plane  in  one  point. 

PROP.  I.     THEOR. 

One  part  of  a  straight  line  cannot  he  in  a  plane  and  another  part  above  it. 

If  it  be  possible  let  AB,  part  of  the  straight  line  ABC,  be  in  the  plane, 
and  the  part  BC  above  it :  and  since  the 

straight  line  AB  is  in  the  plane,  it  can  be  Q, 

produced  in  that  plane  (2.  Post.  1.) ;  let 
it  be  produced  to  D  :    Then  ABC  and  y 

ABD  are  two   straight  lines,  and  they  \  ^ 

have  the  common  segment  AB,  which  is  \    r =d^ 

impossible  (Cor.  def.  3.  1.).     Therefore  \_____- 

ABC  is  not  a  straight  line. 


A 


PROP.  IL    THEOR 

Any  three  straight  lines  which  meet  one  another^  not  in  the  samepointf  are  in 

one  plane. 

Let  the  three  straight  lines  AB,  CD,  CB  meet  one  another  in  the  points 
B,  C  and  E  ;  AB,  CD,  CB  are  in  one  plane. 

Let  any  plane  pass  through  the  straight  line 
EB,  and  let  the  plane  be  turned  about  EB,  pro- 
duced, if  necessary,  until  it  pass  through  the 
point  C  :  Then,  because  the  points  E,  C  are  in 
this  plane,  the  straight  line  EC  is  in  it  (def.  5.1.): 
for  the  same  reason,  the  straight  line  BC  is  in 
the  same  ;  and,  by  the  hypothesis,  EB  is  in  it ; 
therefore  the  three  straight  lines  EC,  CB,  BE 
are  in  one  plane :  but  the  whole  of  the  lines  DC, 
AB,  and  BC  produced,  are  in  the  same  plane 
with  the  parts  of  them  EC,  EB,  BC  (1.  2. 
Sup.).  Therefore  AB,  CD,  CB,  are  all  in  one 
plane. 

CoR.     It  is  manifest,  that  any  two  straight  lines  which  cut  one  another 
are  in  one  plane  :  Also,  that  any  three  poihts  whatever  are  in  one  plane 


OF  GEOMETRY.    BOOK  II. 


183 


PROP.  III.    THEOR. 
If  two  planes  cut  one  another ^  their  common  section  is  a  straight  line. 

Let  two  planes  AB,  BC  cut  one  another, 
and  let  B  and  D  be  two  points  in  the  line  of 
their  common  section.  From  B  to  D  draw  the 
straight  line  BD ;  and  because  the  points  B 
and  I)  are  in  the  plane  AB,  the  straight  line 
BD  is  in  that  plane  (def.  5.  1.) :  for  the  same 
reason  it  is  in  the  plane  CB  ;  the  straight  line 
BD  is  therefore  common  to  the  planes  AB 
and  BC,  or  it  is  the  common  section  of  these 
planes. 


PROP.  IV.    THEOR. 

If  a  straight  line  stand  at  right  angles  to  each  of  two  straight  lines  in  the 
point  of  their  intersection^  it  will  also  he  at  right  angles  to  the  plane  in 
which  these  lines  are. 

Let  the  straight  line  AB  stand  at  right  angles  to  each  of  the  straight 
lines  EF,  CD  in  A,  the  point  of  their  intersection :  AB  is  also  at  right  an- 
gles to  the  plane  passing  through  EF,  CD. 

Through  A  draw  any  line  AG  in  the 
plane  in  which  are  EF  and  CD  ;  let  G  be 
any  point  in  that  lino  ;  draw  GH  parallel 
to  AD  ;  and  make  HF=HA,  join  FG ;  and 
when  produced  let  it  meet  CA  in  D ;  join 
BD,  BG,  BF.  Because  GH  is  parallel  to 
AD,  and  FH=HA  :  therefore  FG=GD, 
so  that  the  line  DF  is  bisected  in  G.  And 
because  BAD  is  a  right  angle,  BD^srAB^ 
4-AD2  (47.  L);  and  for  the  same  reason, 
BF2  =  AB2+AF2,  therefore  BD2+BF2= 
2AB2  +  AD2  +  AF2 ;  and  because  DF  is 
bisected  in  G  (A.  2.),  AD2+ AF2=2AG2+ 

2GF2,  therefore  BD2+BF2=2AB2+2AG2       -^  T> 

+2GF2.  But  BD2  +  BF2=  (A.  2.)  2BG24-2GF2,  therefore  2BG2+ 
2GF2=2AB2+2AG2+2GF2  ;  and  taking  2GF2  from  both,  2BG2=:2AB2 
4-2AG2,  or  BG2=AB2+AG2 ;  whence  BAG  (48.  1.)  is  a  right  angle. 
Now  AG  is  any  straight  line  drawn  in  the  plane  of  the  lines  AD,  AF  ;  and 
when  a  straight  line  is  at  right  angles  to  any  straight  line  which  it  meets 
with  in  a  plane,  it  is  at  right  angles  to  the  plane  itself  (def.  1 .  2.  Sup.).  AB 
is  therefore  at  right  angles  to  the  plane  of  the  lines  AF,  AD. 


184 


SUPPLEMENT  TO  THE  ELEMENTS 


PROP.  V.    THEOR. 

If  three  straight  lines  meet  all  in  one  pointy  and  a  straight  line  stand  at  right 
angles  to  each  of  them  in  that  point ;  these  three  straight  lines  are  in  one 
and  the  sam^e  plane. 

Let  the  straight  line  AB  stand  at  right  angles  to  each  of  the  straight 
lines  BO,  BD,  BE,  in  B,  the  point  where  they  meet ;  BC,  BD,  BE  are  in 
one  and  the  same  plane. 

If  not,  let  BD  and  BE,  if  possible,  be  in  one  plane,  and  BC  be  above  it ; 
and  let  a  plane  pass  through  AB,  EC,  the  common  section  of  which  with 
the  plane,  in  which  BD  and  BE  are,  shall  be  a  straight  (3.  2.  Sup.)  line  ; 
let  this  be  BF :  therefore  the  three  straight  lines  AB,  BC,  BF  are  all  in 
one  plane,  viz.  that  which  passes  through  AD,  BC ;  and  because  AB 
stands  at  right  angles  to  each  of  the  straight  lines  BD,  BE,  it  is  also  at 
right  angles  (4.  2.  Sup.)  to  the  plane  passing 
through  them ;  and  therefore  makes  right  an- 
gles with  every  straight  line  meeting  it  in  that 
plane  ;  but  BF  which  is  in  that  plane  meets  it ; 
therefore  the  angle  ABF  is  a  right  angle ;  but 
the  angle  ABC,  by  the  hypothesis  is  also  a  right 
angle ;  therefore  the  angle  ABF  is  equal  to  the 
angle  ABC,  and  they  are  both  in  the  same 
plane,  which  is  impossible :  therefore  the  straight 
line  BC  is  not  above  the  plane  in  which  are  BD 
and  BE  :  Wherefore  the  three  straight  lines 
BC,  BDj'BE  are  in  one  and  the  same  plane. 


PROP.  VL    TPIEOR. 

Two  straight  lines  which  are  at  right  angles  to  the  same  plane ^  are  parallel  to 

one  another, 

Let*the  straight  lines  AB,  CD  be  at  right  angles  to  the  same  plane  BDE  ; 
AB  is  parallel  to  CD. 

Let  them  meet  the  plane  in  the  points  B,  D. 
Draw  DE  at  right  angles  to  DB,  in  the  plane  BDE, 
and  let  E  be  any  point  in  it :  Join  AE,  AD,  EB. 
Because  ABE  is  a  right  angle,  AB2+BE2=  (47. 1.) 
AE2,  and  because  BDE  is  a  right  angle,  BE2=BD2 
+  DE2;  therefore  AB2-f  BD2+DE2=AE2  ;  now, 
AB2-f.BD2=AD2,  because  ABD  is  a  right  angle, 
therefore  AD2+DE2=AE2,  and  ADE  is  therefore 
a  (48.  1.)  right  angle.  Therefore  ED  is  perpendi- 
cular to  the  three  lines  BD,  DA,  DC,  whence  these 
lines  are  in  one  plane  (5.  2.  Sup.).  But  AB  is  in  the 
plane  in  which  are  BD,  DA,  because  any  three 
straight  lines,  which  meet  one  another,  are  in  one 


OF  GEOMETRY.    BOOK  II.  185 

plane  (2. 2.  Sup.)  :  therefore  AB,  BD,  DC  are  in  one  plane  ;  and  each  of 
the  angles  ABD,  BDC  is  a  right  angle  ;  therefore  AB  is  parallel  (Cor.  28. 
l.)toCD. 

PROP.  VII.    THEOR. 

Jftwo  straight  lines  he  parallel,  and  one  of  them  at  right  angles  to  a  plane  ; 
the  other  is  also  at  right  angles  to  the  sarne  plane. 

Let  AB,  CD  be  two  parallel  straight 
lines,  and  let  one  of  them  AB  be  at 
right  angles  to  a  plane  ;  the  other  CD 
is  at  right  angles  to  the  same  plane. 

For,  if  CD  be  not  perpendicular  to 
the  plane  to  which  AB  is  perpendicular, 
let  DG  be  perpendicular  to  it.  Then 
(6.  2.  Sup.)  DG  is  parallel  to  AB  :  DG 
and  DC  therefore  are  both  parallel  to 
AB,  and  are  drawn  through  the  same 
point  D,  which  is  impossible  (11.  Ax. 
1.). 

PROP.  VIII.     THEOR. 

Two  straight  lines  which  are  each  of  them  parallel  to  the  same  straight  line, 
though  not  both  in  the  same  plane  with  it,  are  parallel  to  one  another. 

Let  AB,  CD  be  each  of  them  parallel  to  EF,  and  not  in  the  same  plane 
with  it ;  AB  shall  be  parallel  to  CD. 

In  EF  take  any  point  G,  from  which  draw,  in  the  plane  passing  through 
EF,  AB,  the  straight  line  GH  at  right  angles  to  EF ;  and  in  the  plane 
passing  through  EF,  CD,  draw  GK  at  right  angles  to  the  same  EF. 
And  because  EF  is  perpendicular  both  to  GH  and  GK,  it  is  perpendicular 
(4.  2.  Sup.)  to  the  plane  HGK  passing  through  them :  and  E^  is  parallel 
to  AB  ;  therefore  AB  is  at  right 
angles  (7.  2.  Sup.)  to  the  plane 
HGK.  For  the  same  reason,  CD 
is  likewise  at  right  angles  to  the 
plane  HGK.  Therefore  AB,  CD 
are  each  of  them  at  right  angles 
to  the  plane  HGK.  But  if  two 
straight  lines  are  at  right   angles  ^ 

to  the  same  plane,  they  are  paral-  -K 

lei  (6.  2.  Sup.)  to  one  another.     Therefore  AB  is  parallel  to  CD. 

PROP.  IX.     THEOR. 

If  two  straight  lines  meeting  one  another  he  parallel  to  two  others  that  meet  one 
another,  though  not  in  the  same  plane  with  the  first  two  ;  the  first  two  and  the 
other  two  shall  contain  equal  angles. 

Let  the  two  straight  lines  AB,  BC  which  meet  one  another,  be  parallel 

24 


186 


SUPPLEMENT  TO  THE  ELEMENTS 


to  the  two  straight  lines  DE,  EF  that  meet  one  another,  and  are  not  in  the 
same  plane  with  AB,  BC.     The  angle  ABC  is  equal  to  the  angle  DEF. 

Take   BA,  BC,  ED,  EF  all  equal  to  one  an- 
other ;  and  join  AD,  CF,  BE,  AC,  DF :  Because 
BA  is  equal  and  parallel  to  ED,  therefore  AD  is 
(33.  1.)  both  equal  and  parallel  to  BE.     For  the 
same   reason,  CF  is  equal  and  parallel  to  BE. 
Therefore  AD  and  CF  are  each  of  them  equal  and 
parallel  to  BE.     But  straight  lines  that  are  paral- 
lel to  the  same  straight  line,  though  not  in  the 
same  plane  with  it,  are  parallel  (8.  2.  Sup.)  to  one 
another.    Therefore  AD  is  parallel  to  CF  ;  and  it 
is  equal  to  it,  and  AC,  DF  join  them  towards  the 
same  parts;  and  therefore  (33.  L)  AC  is  equal 
and  parallel  to  DF.     And  because  AB,  BC  are 
equal  to  DE,  EF,  and  the  base  AC  to  the  base 
DF ;  the  angle  ABC  is  equal  (8.  L)  to  the  angle 
DEF. 


PROP.  X.    PROB. 


To  draw  a  straight  line  perpendicular  to  a  plane,  from  a  given  point  above  it. 

Let  A  be  the  given  point  above  the  plane  BH,  it  is  required  to  draw  from 
the  point  A  a  straight  line  perpendicular  to  the  plane  BH. 

In  the  plane  draw  any  straight  line  BC,  and  from  the  point  A  draw  (Prop. 
12. 1.)  ADpeipendicular  to  BC.  If  then  AD  be  also  perpendicular  to  the 
plane  BH,  the  thing  required  is  already  done  ;  but  if  it  be  not,  from  the 
point  D  draw  (Prop.  ILL),  in  the 
plane  BH,  the  straight  line  DE  at 
right  angles  to  BC  ;  and  from  the 
point  A  draw  AF  perpendicular  to 
DE  ;  and  through  F  draw  (Prop.  31 
L)  GH  parallel  to  BC  :  and  because 
BC  is  at  right  angles  to  ED,  and  DA, 
BC  is  at  right  angles  (4.  2.  Sup.)  to 
the  plane  passing  through  ED,  DA. 
And  GH  is  parallel  to  BC  ;  but  if  two 
straight  lines  be  parallel,  one  of  which  is  at  right  angles  to  a  plane,  the 
other  shall  be  at  right  (7.  2.  Sup.)  angles  to  the  same  plane  ;  wherefore 
GH  IS  at  right  angles  to  the  plane  through  ED,  DA,  and  is  perpendicular 
(def.  1.  2.  Sup.)  to  every  straight  line  meeting  it  in  that  plane.  But  AF, 
which  IS  in  the  plane  through  ED,  DA,  meets  it :  Therefore  GH  is  per- 
pendicular to  AF,  and  consequently  AF  is  perpendicular  to  GH  ;  and  AF 
is  also  perpendicular  to  DE  :  Therefore  AF  is  perpendicular  to  each  of  the 
straight  lines  GH,  DE.  But  if  a  straight  line  stands  at  right  angles  to 
each  of  two  straight  lines  in  the  point  of  their  intersection,  it  is  also  at  right 
angles  to  the  plane  passing  through  them  (4.  2.  Sup.).  And  the  plane 
passmg  through  ED,  GH  is  the  plane  BH  ;  therefore  AF  is  perpendicular 


OF  GEOMETRY.    BOOK  II. 


187 


to  the  plane  BH ;  so  that,  from  the  given  point  A,  above  the  plane  BH, 
the  straight  line  AF  is  drawn  perpendicular  to  that  plane. 

CoR.  If  it  be  required  from  a  point  C  in  a  plane  to  erect  a  perpen- 
dicular to  that  plane,  take  a  point  A  above  the  plane,  and  draw  AF  per- 
pendicular to  the  plane ;  then,  if  from  0  a  line  be  drawn  parallel  to  AF, 
it  will  be  the  perpendicular  required ;  for  being  parallel  to  AF  it  will  be 
perpendicular  to  the  same  plane  to  which  AF  is  perpendicular  (7.  2.  Sup.). 


c 


PROP.  XI.    THEOR. 

From  the  same  point  in  a  plane,  there  cannot  he  two  straight  lines  at  right 
angles  to  the  plane,  upon  the  same  side  of  it ;  And  there  can  he  hut  one 
perpendicular  to  a  plane  from  a  point  ahove  it. 

For  if  it  be  possible,  let  the  two  straight  lines  AC,  AB  be  at  right  angles 
to  a  given  plane  from  the  same  point  A  in  the  plane,  and  upon  the  same 
side  of  it ;  and  let  a  plane  pass  through  BA,  AC  ;  the  common  section  of 
this  plane  with  the  given  plane  is  a  straight  (3.  2.  Sup.)  line  passing  through 
A  :  Let  DAE  be  their  common  section  :  Therefore  the  straight  lines  AB, 
AC,  DAE  are  in  one  plane :  And  because  CA  is  at  right  angles  to  the 
given  plane,  it  makes  right  angles  with  every 
straight  line  meeting  it  in  that  plane.  But 
DAE,  which  is  in  that  plane,  meets  CA :  there- 
fore CAE  is  a  right  angle.  For  the  same  rea- 
son BAE  is  a  right  angle.  Wherefore  the  an- 
gle CAE  is  equal  to  the  angle  BAE  ;  and 
they  are  in  one  plane,  which  is  impossible. 

Also,  from  a  point  above  a  plane,  there  can  be.      =—  — - 

but  one  perpendicular  to  that  plane  ;  for  if  there       -^  -^  -^ 

could  be  two,  they  would  be  parallel  (6.  2.  Sup.)  to  one  another,  which  is 
absurd. 

PROP.  XII.     THEOR. 

Planes  to  which  the  same  straight  line  is  perpendicular,  are  parallel  to  one 

another. 

Let  the  straight  line  AB  be  perpendicular  to 
each  of  the  planes  CD,  EF  :  these  planes  are  pa- 
rallel to  one  another. 

If  not,  they  must  meet  one  another  when  pro- 
duced, and  their  common  section  must  be  a  straight 
line  GH,  in  which  take  any  point  K,  and  join  AK, 
BK  :  Then,  because  AB  is  perpendicular  to  the 
plane  EF,  it  is  perpendicular  (def.  1.  2.  Sup.)  to 
the  straight  line  BK  which  is  in  that  plane,  and 
therefore  ABK  is  a  right  angle.  For  the  same 
reason,  BAK  is  a  right  angle  ;  wherefore  the  two 
angles  ABK,  BAK  of  the  triangle  ABK  are 
equal  to  two  right  angles,  which  is  impossible, 


188 


SUPPLEMENT  TO  THE  ELEMENTS 


(17.  1.):  Therefore  the  planes  CD,  EF,  though  produced,  do  not  meet 
one  another  j  that  is,  they  are  parallel  (def.  7.  2.  Sup.). 


PROP.  XIII.     THEOR. 

If  two  straight  lines  meeting  one  another,  he  parallel  to  two  straight  lines 
which  also  meet  one  another,  but  are  not  in  the  same  plane  with  the  first 
two :  the  plane  which  passes  through  the  first  two  is  parallel  to  the  plane 
passing  through  the  others. 

Let  AB,  BC,  two  straight  lines  meeting  one  another,  be  parallel  to  DE, 
EF  that  meet  one  another,  but  are  not  in  the  same  plane  with  AB,  BC  : 
The  planes  through  AB,  BC,  and  DE,  EF  shall  not  meet,  though  pro- 
duced. 

From  the  point  B  draw  BG  perpendicular  (10.  2.  Sup.)  to  the  plane 
which  passes  through  DE,  EF,  and  let  it  meet  that  plane  in  G ;  and 
through  G  draw  GH  parallel  to  ED  (Prop.  31.  1.),  and  GK  parallel  to  EF  : 
And  because  BG  is  perpendicular  to  the  plane  through  DE,  EF,  it  must 
make  right  angles  with  every 
straight  line  meeting  it  in  that 
plane  (1.  def.  2.  Sup.).  But 
the  straight  lines  GH,  GK  in 
that  plane  meet  it :  Therefore 
each  of  the  angles  BGH,  BGK 
is  a  right  angle  :  And  because 
BA  is  parallel  (8.  2.  Sup.)  to 
GH  (for  each  of  them  is  paral- 
lel to  DE),  the  angles  GBA, 
BGH  are  together  equal  (29. 
1.)  to  two  right  angles:  And 

BGH  is  a  right  angle ;  therefore  also  GBA  is  a  right  angle,  and  GB  per- 
pendicular to  BA :  For  the  same  reason,  GB  is  perpendicular  to  BC  : 
Since,  therefore,  the  straight  line  GB  stands  at  right  angles  to  the  two 
straight  lines  BA,  BC,  that  cut  one  another  in  B ;  GB  is  perpendicular 
(4-  2.  Sup.)  to  the  plane  through  BA,  BC  :  And  it  is  perpendicular  to  the 
plane  through  DE,  EF ;  therefore  BG  is  perpendicular  to  each  of  the 
planes  through  AB,  BC,  and  DE,  EF  :  But  planes  to  which  the  same 
straight  line  is  perpendicular,  are  parallel  (12.  2.  Sup.)  to  one  another  : 
Therefore  the  plane  through  AB,  BC,  is  parallel  to  the  plane  through 
DE,  EF. 

CoR.  It  follows  from  this  demonstration,  that  if  a  straight  line  meet 
two  parallel,  planes,  and  be  perpendicular  to  one  of  them,  it  must  be  per- 
pendicular to  the  other  also. 


OF  GEOMETRY.    BOOK  II. 


189 


PROP.  XIV.    THEOR. 

Jftwo  parallel  planes  be  cut  hy  another  plane,  their  common  sections  with  it 

are  parallels. 


Let  the  parallel  planes  AB, 
CD,  be  cut  by  the  plane  EFHG, 
and  let  their  common  sections  with 
it  be  EF,  GH ;  EF  is  parallel  to 
GH. 

For  the  straight  lines  EF  and 
GH  are  in  the  same  plane,  viz. 
EFHG  which  cuts  the  planes 
AB  and  CD ;  and  they  do  not 
meet  though  produced;  for  the 
planes  in  which  they  are  do  not 


N^ 


A>\ 


meet;  therefore  EF  and  GH  are  parallel  (def.  30.  1.). 


PROP.  XV.    THEOR. 

If  two  parallel  planes  he  cut  by  a  third  plane,  they  have  the  same  inclination 

to  that  plane. 

Let  AB  and  CD  be  two  parallel  planes,  and  EH  a  third  plane  cutting 
them ;  The  planes  AB  and  CD  are  equally  inclined  to  EH. 

Let  the  straight  lines  EF  and  GH  be  the  common  section  of  the  plane 
EH  with  the  two  planes  AB  and  CD  ;  and  from  K,  any  point  in  EF,  draw 
in  the  plane  EH  the  straight  line  KM  at  right  angles  to  EF,  and  let  it 
meet  GH  in  L ;  draw  also  KN  at  right  angles  to  EF  in  the  plane  AB : 
and  through  the  straight  lines  KM,  KN,  let  a  plane  be  made  to  pass,  cut- 
ting the  plane  CD  in  the  line  LO.  And  because  EF  and  GH  are  the 
common  sections  of  the  plane  EH  with  the  two  parallel  planes  AB  and 
CD,  EF  is  parallel  to  GH  (14.  2.  Sup.).  But  EF  is  at  right  angles  to 
the  plane  that  passes  through  KN  and  KM  (4.  2.  Sup.),  because  it  is  at 
right  angles  to  the  lines  KM  and  KN  :  therefore  GH  is  also  at  right  an- 
gles to  the  same  plane  (7.  2.  Sup.),  and  it  is  therefore  at  right  angles  to 


K^ 


5  D 


J3C: 


o 


M 


190 


SUPPLEMENT  TO  THE  ELEMENTS 


ihe  lines  LM,  LO  which  it  meets  in  that  plane.  Therefore,  since  LM  and 
LO  are  at  right  angles  to  LG,  the  common  section  of  the  two  planes  CD 
and  EH,  the  angle  OLM  is  the  inclination  of  the  plane  CD  to  the  plane 
EH  (4.  def.  2.  Sup.).  For  the  same  reason  the  angle  MKN  is  the  inclina- 
tion of  the  plane  AB  to  the  plane  EH.  But  because  KN  and  LO  are  pa- 
rallel, being  the  common  sections  of  the  parallel  planes  AB  and  CD  with 
a  third  plane,  the  interior  angle  NKM  is  equal  to  the  exterior  angle  OLM 
(29.  1.)  ;  that  is,  the  inclination  of  the  plane  AB  to  the  plane  EH,  is  equal 
to  the  inclination  of  the  plane  CD  to  the  same  plane  EH. 


PROP.  XVL    THEOR. 
If  two  straight  lineshe  cut  hy  parallel  planes,  they  must  he  cut  in  the  same  ratto. 

Let  the  straight  lines  AB,  CD  be  cut  by  the  parallel  planes  GH,  KL, 
MN,  in  the  pomts  A,  E,  B ;  C,  F,  D : 
As  AE  is  to  EB,  so  is  CF  to  FD. 

Join  AC,  BD,  AD,  and  let  AD  meet 
the  plane  KL  in  the  point  X ;  and  join 
EX,  XF :  Because  the  two  parallel 
planes  KL,  MN  are  cut  by  the  plane 
EBDX,  the  common  sections  EX,  BD, 
are  parallel  (14.  2.  Sup.).  For  the  same 
reason,  because  the  two  parallel  planes 
GH,  KL  are  cut  by  the  plane  AXFC, 
the  common  sections  AC,  XF  are  paral- 
lel :  And  because  EX  is  parallel  to  BD, 
a  side  of  the  triangle  ABD,  as  AE  to 
EB,  so  is  ( 2.  6.)  AX  to  XD.  Again,  be- 
cause XF  is  parallel  to  AC,  aside  of  the 
triangle  ADC,  AX  to  XD,  so  is  CF  to 
FD  :  and  it  was  proved  that  AX  is  to  XD, 
as  AE  to  EB  :  Therefore  (11.  5.),  as  AE 
to  EB,  so  is  CF  to  FD. 

PROP.  XVIL     THEOR. 

If  a  straight  line  he  at  right  angles  to  a  plane,  every  plane  which  passes  through 
that  line  is  at  right  angles  to  the  first  mentioned  plane. 

Let  the  straight  line  AB  be  at  right  angles  to  the  plane  CK  ;  every  plane 
which  passes  through  AB  is  at  right  angles  to  the  plane  CK. 

Let  any  plane  DE  pass  through  AB,  and  let  CE  be  the  common  section 
of  the  planes  DE,  CK  ;  take  any  point  F  in  CE,  from  which  draw  FG  in 
the  plane  DE  at  right  angles  to  CE  :  And  because  AB  is  perpendicular 
to  the  plane  CK,  therefore  it  is  also  perpendicular  to  every  straight  line 
meetmgit  in  that  plane  (1.  def.  2.  Sup.);  and  consequently  it  is  perpen- 
dicular to  CE  :  Wherefore  ABF  is  a  right  angle  ;  But  GFB  is  likewise  a 
right  angle  ;  therefore  AB  is  parallel  (28.  1.)  to  FG.  And  AB  is  at  right 
angles  to  the  plane  CK  :  therefore  FG  is  also  at  right  angles  to  the  same 


OF  GEOMETRY.    BOOK  II. 


19] 


plane  (7.  2.  Sup.).  But  one  plane  is 
at  right  angles  to  another  plane  when 
the  straight  lines  drawn  in  one  of  the 
planes,  at  right  angles  to  their  com- 
mon section,  are  also  at  right  angles 
to  the  other  plane  (def.  2. 2.  Sup.) ;  and 
any  straight  line  FG  in  the  plane  DE, 
which  is  at  right  angles  to  CE,  the 
common  section  of  the  planes,  has  been 
proved  to  be  perpendicular  to  the  other 
plane   OK ;  therefore  the  plane  DE 

is  at  right  angles  to  the  plane  CK.  In  like  manner,  it  may  be  proved 
that  all  the  planes  which  pass  through  AB  are  at  right  angles  to  the  plane 
CK. 


D            6       A     }J 

K 

\ 

c 


B      E 


PROP.  XVIII.     THEOR. 

If  two  planes  cutting  one  another  be  each  of  them  perpendicular  to  a  third  plane 
their  common  section  is  perpendicular  to  the  same  plane. 

Let  the  two  planes  AB,  BC  be  each  of  them  perpendicular  to  a  third 
plane,  and  BD  be  the  common  section  of  the  first  two ;  BD  is  perpendicular 


to  the  plane  ADC. 

From  D  in  the  plane  ADC,  draw  DE  perpen- 
dicular to  AD,  and  DF  to  DC.  Because  DE  is 
perpendicular  to  AD,  the  common  section  of  the 
planes  AB  and  ADC ;  and  because  the  plane 
AB  is  at  right  angles  to  ADC,  DE  is  at  right 
angles  to  the  plane  AB  (def.  2. 2.  Sup.),  and  thei;e- 
fore  also  to  the  straight  line  BD  in  that  plane 
(def.  1. 2.  Sup.).  For  the  same  reason,  DF  is  at 
right  angles  to  DB.  Since  BD  is  therefore  at 
right  angles  to  both  the  lines  DE  and  DF,  it  is 
at  right  angles  to  the  plane  in  which  DE  and 
DF  are,  that  is,  to  the  plane  ADC  (4.  2.  Sup.). 


B 


D 


A.E 


E 


PROP.  XIX.    PROB. 

Two  straight  lines  not  in  the  same  plane  being  given  in  position^  to  draw  a 
straight  line  perpendicular  to  them  both. 

Let  AB  and  CD  be  the  given  lines,  which  are  not  in  the  same  plane  ;  it 
is  required  to  draw  a  straight  line  which  shall  be  perpendicular  both  to  AB 
and  CD. 

In  AB  take  any  point  E,  and  through  E  draw  EF  parallel  to  CD,  and 
let  EG  be  drawn  perpendicular  to  the  plane  which  passes  through  EB, 
BF  (10.  2.  Sup.).  Through  AB  and  EG  let  a  plane  pass,  viz.  GK,  and  let 
this  plane  meet  CD  in  H ;  from  H  draw  HK  perpendicular  to  AB  ;  and 
HK  is  the  line  required.     Through  H,  draw  HG  parallel  to  AB. 


192  SUPPLEMENT  TO  THE  ELEMENTS 


Then,  since  HK  and  GE,  which  are  in  the  same  plane,  are  both  at  right 
angles  to  the  straight  line  AB,  they  are  parallel  to  one  another.  And  be- 
cause the  lines  HG,  HD  are  parallel  to  the  lines  EB,  EF,  each  to  each, 
the  plane  GHD  is  parallel  to  the  plane  (13.  2.  Sup.)  BEF ;  and  therefore 
EG,  which  is  perpendicular  to  the  plane  BEF,  is  perpendicular  also  to  the 
plane  (Cor.  13.  2.  Sup.)  GHD.  Therefore  HK,  which  is  parallel  to  GE, 
is  also  perpendicular  to  the  plane  GHD  (7.  2.  Sup.),  and  it  is  therefore  per- 
pendicular to  HD  (def.  1.  2.  Sup.),  which  is  in  that  plane,  and  it  is  also 
perpendicular  to  AB ;  therefore  HK  is  drawn  perpendicular  to  the  two 
given  lines,  AB  and  CD. 

PROP.'  XX.    THEOR. 

If  a  solid  angle  be  cohtained  by  three  plane  angles,  any  two  of  these  angles  are 
greater  than  the  third. 

Let  the  solid  angle  at  A  be  contained  by  the  three  plane  angles  BAG, 
CAD,  DAB.     Any  two  of  them  are  greater  than  the  third. 

If  the  angles  BAG,  CAD,  DAB  be  all  equal,  it  is  evident  that  any  two 
of  them  are  greater  than  the  third.  But  if  they  are  not,  let  BAG  be  that 
angle  which  is  not  less  than  either  of  the  other  two,  and  is  greater  than 
one  of  them,  DAB  ;  and  at  the  point  A  in  the 
straight  line  AB,  make  in  the  plane  which 
passes  through  BA,  AC,  the  angle  BAE  equal 
(Prop. 23.  1.)  to  the  angle  DAB;  and  make 
AE  equal  to  AD,  and  through  E  draw  BEG 
cutting  AB,  AC  in  the  points  B,  C,  and  join 
DB,  DC.  And  because  DA  is  equal  to  AE, 
and  AB  is  common  to  the  two  triangles  ABD, 
ABE,  and  also  the  angle  DAB  equal  to  the 
angle  EAB  ;  therefore  the  base  DB  is  equal  (4.  1.)  to  the  base  BE.  And 
because  BD,  DC  are  greater  (20.  1.)  than  CB,  and  one  of  them  BD  has 
been  proved  equal  to  BE,  a  part  of  CB,  therefore  the  other  DC  is  greater 
than  the  remaining  part  EC.  And  because  DA  is  equal  to  AE,  and  AC 
common,  but  the  base  DC  greater  than  the  base  EC  ;  therefore  the  angle 
DAG  is  greater  (25.  1.)  than  the  angle  EAC  ;  and,  by  the  construction, 


OF  GEOMETRY.    BOOK  II.  193 

the  angle  DAB  is  equal  to  the  angle  BAE  ;  wherefore  the  angles  DAB, 
DAC  are  together  greater  than  BAE,  EAC,  that  is,  than  the  angle  BAG. 
But  BAG  is  not  less  than  either  of  the  angles  DAB,  DAG  ;  therefore 
BAG,  with  either  of  them,  is  greater  than  the  other. 


PROP.  XXI.     THEOR. 

The  plane  angles  which  contain  any  solid  angle  are  together  less  than  four 

right  angles. 

Let  A  be  a  solid  ang  e  contained  by  any  number  of  plane  angles  BAG, 
GAD,  DAE,  EAF,  FAB  ;  these  together  are  less  than  four  right  angles. 

Let  the  planes  which  contain  the  solid  angle  at  A  be  cut  by  another 
plane,  and  let  the  section  of  them  by  that  plane  be  the  rectilineal  figure 
BGDEF.  And  because  the  solid  angle  at  B  is  contained  by  three  plane 
angles  GBA,  ABF,  FBG,  of  which  any  two 
are  greater  (20.  2.  &up.)  than  the  third,  the 
angles  GBA,  ABF  are  greater  than  the  an- 
gle FBG  :  For  the  same  reason,  the  two 
plane  angles  at  each  of  the  points  G,  D,  E, 
F,  viz.  the  angles  which  are  at  the  bases  of 
the  triangles  having  the  common  vertex  A, 
are  greater  than  the  third  angle  at  the  same 
point,  which  is  one  of  the  angles  of  the  figure 
BGDEF  :  therefore  all  the  angles  at  the 
bases  of  the  triangles  are  together  greater 
than  all  the  angles  of  the  figure :  and  be- 
cause all  the  angles  of  the  triangles  are  to- 
gether equal  to  twice  as  many  right  angles  as  there  are  triangles  (32.  1.)  > 
that  is,  as  there  are  sides  in  the  figure  BGDEF  ;  and  because  all  the  an- 
gles of  the  figure,  together  with  four  right  angles,  are  likewise  equal  to 
twice  as  many  right  angles  as  there  are  sidesinthefigure(l  cr.  32.  l.);there- 
fore  all  the  angles  of  the  triangles  are  equal  to  all  the  angles  of  the  rectili- 
neal figure,  together  with  four  right  angles.  But  all  the  angles  at  the  bases 
of  the  triangles  are  greater  than  all  the  angles  of  the  rectilineal,  as  has 
been  proved.  Wherefore,  the  remaining  angles  of  the  triangles,  viz.  those 
at  the  vertex,  which  contain  the  solid  angle  at  A,  are  less  than  four  right 
angles. 

Otherwise. 

Let  the  sum  of  all  the  angles  at  the  bases  of  the  triangles  =S ;  the 
sum  of  all  the  angles  of  the  rectilineal  figure  BGDEF=^ ;  the  sum  of  the 
plane  angles  at  A=X,  and  let  R=  a  right  angle. 

Then,  because  S-j-X=:  twice  (32.  1.)  as  many  right  angles  as  there  are 
triangles,  or  as  there  are  sides  of  the  rectilineal  figure  BGDEF,  and  as 
-^-|-4R  is  also  equal  to  twice  as  many  right  angles  as  there  are  sides  of  the 
same  figure  ;  therefore  S4-X=2'-f-4R.  But  because  of  the  three  plane 
angles  which  contain  a  solid  angle,  any  two  are  greater  than  the  third, 

25 


194  SUPPLEMENT  TO  THE  ELEMENTS,  &c. 

S/^  ;  and  therefore  X/4R  ;  that  is,  the  sum  of  the  plane  angles  which' 
contain  the  solid  angle  at  A  is  less  than  four  right  angles. 

SCHOLIUM.  "" 

It  is  evident,  that  when  any  of  the  angles  of  the  figure  BCDEF  is  ex- 
terior, like  the  angle  at  D,  in  the  an- 
nexed figure,  the  reasoning  in  the  ^ 
above  proposition  does  not  hold,  be- 
cause the  solid  angles  at  the  base 
are  not  all  contained  by  plane  an- 
gles, of  which  two  belong  to  the  tri- 
angidar  planes,  having  their  com- 
mon vertex  in  A,  and  the  third  is  an 
interior  angle  of  the  rectilineal  figure, 

or  base.     Therefore,  it  cannot   be        ^^  /^ 

concluded  that  S  is  necessarily  great-        -O  v^ 

er  than  2.     This  proposition,  therefore,  is  subject  to  a  limitation,  which  is 
farther  explained  in  the  notes  on  this  Book. 


ELEMENTS 

OF 


GEOMETRY 


SUPPLEMENT. 

BOOK  III. 

OF  THE  COMPARISON  OF  SOLIDS. 


DEFINITIONS. 

1.  A  Solid  is  that  which  has  length,  breadth,  and  thickness. 

2.  Similar  solid  figures  are  such  as  are  contained  by  the  same  number  of 
similar  planes  similarly  situated,  and  having  like  inclinations  to  one  an- 
other. 

3.  A  pyramid  is  a  solid  figure  contained  by  planes  that  are  constituted  be- 
twixt one  plane  and  a  point  above  it  in  which  they  meet. 

4.  A  prism  is  a  solid  figure  contained  by  plane  figures,  of  which  two  that 
are  opposite  are  equal,  similar,  and  parallel  to  one  another ;  and  the 
others  are  parallelograms. 

5.  A  parallelepiped  is  a  solid  figure  contained  by  six  quadrilateral  figures, 
whereof  every  opposite  two  are  parallel. 

6.  A  cube  is  a  solid  figure  contained  by  six  equal  squares. 

7.  A  sphere  is  a  solid  figure  described  by  the  revolution  of  a  semicircle 
about  a  diameter,  which  remains  unmoved. 

8.  The  axis  of  a  sphere  is  the  fixed  straight  line  about  which  the  semi- 
circle revolves. 

9.  The  centre  of  a  sphere  is  the  same  with  that  of  the  semicircle. 

10.  The  diameter  of  a  sphere  is  any  straight  line  which  passes  through 
the  centre,  and  is  terminated  both  ways  by  the  superficies  of  the  sphere. 


196 


SUPPLEMENT  TO  THE  ELEMENTS 


IL  A  cone  is  a  solid  figure  described  by  tbe  revolution  of  a  right  angled 
triangle  about  one  of  the  sides  containing  the  right  angle,  which  side 
remains  fixed. 

12.  The  axis  of  a  cone  is  the  fixed  straight  line  about  which  the  triangle 
revolves. 

13.  The  base  of  a  cone  is  the  circle  described  by  that  side,  containing  the 
right  angle,  which  revolves. 

14.  A  cylinder  is  a  solid  figure  described  by  the  revolution  of  a  right  an- 
gled parallelogram  about  one  of  its  sides,  which  remains  fixed. 

15.  The  axis  of  a  cylinder  is  the  fixed  straight  line  about  which  the  paral- 
lelogram revolves. 

16.  The  bases  of  a  cylinder  are  the  circles  described  by  the  two  revolving 
opposite  sides  of  the  parallelogram. 

17.  Similar  cones  and  cylinders  are  those  which  have  their  axes,  and  the 
diameters  of  their  bases  proportionals. 

PROP.  I.     THEOR. 

If  two  solids  he  contained  hy  the  same  number  of  equal  and  similar  planes 
sindlarly  situated^  and  if  the  inclination  of  any  two  contiguous  planes  in  the 
one  solid  be  the  same  with  the  inclination  of  the  two  equals  and  similarly 
situated  planes  in  the  other ^  the  solids  themselves  are  equal  and  similar. 

Let  AG  and  KQ  be  two  solids  contained  by  the  same  number  of  equal 
and  similar  planes,  similarly  situated  so  that  the  plane  AC  is  similar  and 
equal  to  the  plane  KM,  the  plane  AF  to  the  plane  KP  ;  BG  to  LQ,  GD 
to  QN,  DE  to  NO,  and  FH  to  PR.  Let  also  the  inclination  of  the  plane 
AF  to  the  plane  AC  be  the  same  with  that  of  the  plane  KP  to  the  plane 
KM,  and  so  of  the  rest ;  the  solid  KQ  is  equal  and  similar  to  the  solid  AG. 

Let  the  solid  KQ  be  applied  to  the  solid  AG,  so  that  the  bases  KM  and 


II 


3J 


M 


AC,  which  are  equal  and  similar,  may  coincide  (8.  Ax.  1.),  the  point  N 
coinciding  with  the  point  D,  K  with  A,  L  with  B,  and  so  on.  And  be- 
cause the  plane  KM  coincides  with  the  plane  AC,  and,  by  hypothesis,  the 


OF  GEOMETRY.     BOOK  III.  197 

inclination  of  KR  to  KM  is  the  same  with  the  inclination  of  AH  to  AC, 
the  plane  KR  will  be  upon  the  plane  AH,  and  will  coincide  with  it,  because 
they  are  similar  and  equal  (8.  Ax.  1.),  and  because  their  equal  sides  KN 
and  AD  coincide.  And  in  the  same  manner  it  is  shewn  that  the  other 
planes  of  the  solid  KQ  coincide  with  the  other  planes  of  the  solid  AG, 
each  with  each :  wherefore  the  solids  KQ  and  AG  do  wholly  coincide, 
and  are  equal  and  similar  to  one  another. 

PROP.  n.     THEOR. 

Jf  a  solid  be  contained  by  sixplanes,  two  and  two  of  which  areparallel^  the  op- 
posite  planes  are  similar  and  equal  parallelograms. 

Let  the  solid  CDGH  be  contained  by  the  parallel  planes  AC,  GF  ;  BG, 
CE  ;  FB,  AE  :  its  opposite  planes  are  similar  and  equal  parallelograms. 

Because  the  two  parallel  planes  BG,  CE,  are  cut  by  the  plane  AC,  their 
common  sections  AB,  CD  are  parallel  (14.  2.  Sup.).  Again,  because  the 
two  parallel  planes  BF,  AE  are  cut  by  the  plane  AC,  their  common  sec- 
tions AD,  BC  are  parallel  (14.  2.  Sup.) :  and  AB  is  parallel  to  CD  ;  there- 
fore AC  is  a  parallelogram.  In  like  manner,  it  may  be  proved  that  each 
of  the  figures  CE,  FG,  GB,  BE,  AE  is  a  pa- 
rallelogram ;  join  AH,  DF ;  and  because  AB 
is  parallel  to  DC,  and  BH  to  CF ;  the  two 
straight  lines  AB,  BH,  which  meet  one  an- 
other, are  parallel  to  DC  and  CF,  which  meet 
one  another ;  wherefore,  though  the  first  two 
are  not  in  the  same  plane  with  the  other  two, 
they  contain  equal  angles  (9.  2.  Sup.) ;  the 
angle  ABH  is  therefore  equal  to  the  angle 
DCF.  And  because  AB,  BH,  are  equal  to  DC,  CF,  and  the  angle  ABH 
equal  to  the  angle  DCF  ;  therefore  the  base  AH  is  equal  (4.  1.)  to  the  base 
DF,  and  the  triangle  ABH  to  the  triangle  DCF  :  For  the  same  reason, 
the  triangle  AGH  is  equal  to  the  triangle  DEF  :  and  therefore  the  paral- 
lelogram BG  is  equal  and  similar  to  the  parallelogram  CE.  In  the  same 
manner,  it  may  be  proved,  that  the  parallelogram  AC  is  equal  and  similar 
to  the  parallelogram  GF,  and  the  parallelogram  AE  to  BF. 

PROP.  III.     THEOR. 

Jfa  solid  parallelopiped  be  cut  by  a  plane  parallel  to  two  of  its  opposite  planes, 
it  will  be  divided  into  two  solids,  which  will  be  to  one  another  as  the  bases. 

Let  the  solid  parallelopiped  ABCD  be  cut  by  the  plane  EV,  which  is 
parallel  to  the  opposite  planes  AR,  HD,  and  divides  the  whole  into  the 
solids  ABFV,  EGCD  :  as  the  base  AEFY  to  the  base  EHCF,  so  is  the 
solid  ABFV  to  the  solid  EGCD. 

Produce  AH  both  ways,  and  take  any  number  of  straight  lines  HM, 
MN,  each  equal  to  EH,  and  any  number  AK,  KL  each  equal  to  EA,  and 
complete  the  parallelograms  LO,  KY,  HQ,  MS,  and  the  solids  LP.  KR, 


158 


SUPPLEMENT  TO  THE  ELEMENTS 


HU,  MT ;  then,  because  the  straight  lines  LK,  KA,  AE  are  all  equal,  and 
also  the  straight  lines  KO,  AY,  EF  which  make  equal  angles  with  LK, 
KA,  AE,  the  parallelograms  LO,  KY,  AF  are  equal  and  similar  (36. 1. 
&  def.  1.6.):  and  likewise    the  parallelograms  KX,  KB,  AG ;  as  also 


X 


B     a 


K 

\ 

P 

\R  K^KdIXtt  \ 

z 

K 

A. 

E 

H 

M 

NT 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

o 


E     O     Q     S 


(2.  3.  Sup.)  the  parallelograms  LZ,  KP,  AR,  because  they  are  opposite 
planes.  For  the  same  reason,  the  parallelograms  EC,  HQ,  MS  are  equal 
(36.  1.  &  def.  I.  6.);  and  the  parallelograms  HG,  HI,  IN,  as  also  (2.  3. 
Sup.)  HD,  MU,  NT  ;  therefore  three  planes  of  the  solid  LP,  are  equal  and 
similar  to  three  planes  of  the  solid  KR,  as  also  to  three  planes  of  the  solid 
AV  :  but  the  three  planes  opposite  to  these  three  are  equal  and  similar  to 
them  (2.  3.  Sup.)  in  the  several  solids ;  therefore  the  solids  LP,  KR,  AV 
are  contained  by  equal  and  similar  planes.  And  because  the  planes  LZ, 
KP,  AR  are  parallel,  and  are  cut  by  the  plane  XV,  the  inclination  of  LZ 
to  XP  is  equal  to  that  of  KP  to  PB  ;  or  of  AR  to  BV  (15.  2.  Sup.)  and 
the  same  is  true  of  the  other  contiguous  planes,  therefore  the  solids  LP, 
KR,  and  AV,  are  equal  to  one  another  (1.  3.  Sup.).  For  the  same  rea- 
son, the  three  solids,  ED,  HU,  MT  are  equal  to  one  another;  therefore 
what  multiple  soever  the  base  LF  is  of  the  base  AF,  the  same  multiple  is 
the  solid  LV  of  the  solid  AV;  for  the  same  reason,  whatever  multiple  the 
base  NF  is  of  the  base  HF,  the  same  multiple  is  the  solid  NV  of  the  solid 
ED  :  And  if  the  base  LF  be  equal  to  the  base  NF,  the  solid  LV  is  equal 
(1.  3.  Sup.)  to  the  solid  NV  ;  and  if  the  base  LF  be  greater  than  the  base 
NF,  the  solid  LV  is  greater  than  the  solid  NV  :  and  if  less,  less.  Since 
then  there  are  four  magnitudes,  viz.  the  two  bases  AF,  FH,  and  the  two 
solids  AV,  ED,  and  of  the  base  AF  and  solid  AV,  the  base  LF  and  solid 
LV  are  any  equimultiples  whatever ;  and  of  the  base  FH  and  solid  ED, 
the  base  FN  and  solid  NV  are  any  equimultiples  whatever ;  and  it  has 
been  proved,  that  if  the  base  LF  is  greater  than  the  base  FN,  the  solid  LV 
is  greater  than  the  solid  NV  ;  and  if  equal,  equal :  and  if  less,  less :  There- 
fore (def.  5.  5.)  as  the  base  AF  is  to  the*  base  FH,  so  is  the  solid  AV  to 
the  solid  ED. 


Cor.     Because  the  parallelogram  AF  is  to  the  parallelogram  FH  as  YF 
to  FC  (1.  6.),  therefore  the  solid  AV  is  to  the  solid  ED  as  YF  to  FG 


OF  GEOMETRY.    BOOK  III.  199 


PROP.  IV.     THEOR. 


/ 

/ 

"\ 

z 

D 

JE" 

/\/ 

33 


If  a  solid  parallelopiped  he  cut  hy  a  plane  passing  through  the  diagonals  of 
two  of  the  opposite  planes,  it  will  he  cut  into  two  equal  prisms. 

Let  AB  be  a  solid  parallelopiped,  and  DE,  OF  the  diagonals  of  the  op- 
posite parallelograms  AH,  GB,  viz.  those  which  are  drawn  betwixt  the 
equal  angles  in  each ;  and  because  CD,  FE  are  each  of  them  parallel  to 
GA,  though  not  in  the  same  plane  with  it,  CD,  FE  are  parallel  (8.  2.  Sup.) ; 
wherefore  the  diagonals  CF,  DE  are  in  the  plane  in  which  the  parallels 
are,  and  are  themselves  parallels  (14.  2.  Sup.) ;  ^  q 

the  plane  CDEF  cuts  the  solid  AB  into  two 
equal  parts. 

Because  the  triangle  CGF  is  equal  (34.  1.)  > 
to  the  triangle  CBF,  and  the  triangle  DAE  to  ^ 
DHE  ;  and  since  the  parallelogram  CA  is  equal 
(2.  3.  Sup.)  and  similar  to  the  opposite  one  BE  ; 
and  the  parallelogram  GE  to  CH  :  therefore  the 
planes  which  contain  the  prisms  CAE,  CBE, 
are  equal  and  similar,  each  to  each  ;  and  they 
are  also  equally  inclined  to  one  another,  because 
the  planes  AC,  EB  are  parallel,  as  also  AF  and 
BD,  and  they  are  cut  by  the  plane  CE  (15.  2.  Sup.).  Therefore  the  prism 
CAE  is  equal  to  the  prism  CBE  (1.3.  Sup.),  and  the  solid  AB  is  cut  into 
two  equal  prisms  by  the  plane  CDEF. 

N.  B.  The  insisting  straight  lines  of  a  parallelopiped,  mentioned  in 
the  following  propositions,  are  the  sides  of  the  parallelograms  betwixt  the 
base  and  the  plane  parallel  to  it. 

PROP.  V.    THEOR. 

Solid  parallelopipeds  upon  the  same  hase,  and  of  the  same  altitude^  the  in- 
sisting  straight  lines  of  which  are  terminated  in  the  same  straight  lines  in 
the  plane  opposite  to  the  hase  are  equal  to  one  another. 

Let  the  solid  parallelopipeds  AH,  AK  be  upon  the  same  base  AB,  and 
of  the  same  altitude,  and  let  their  insisting  straight  lines  AF,  AG,  LM,  LN 
be  terminated  in  the  same  straight  line  FN,  and  let  the  insisting  lines  CD, 
CE,  BH,  BK  be  terminated  in  the  same  straight  line  DK  ;  the  solid  AH 
is  equal  to  the  solid  AK. 

Because  CH,  CK  are  parallelograms,  CB  is  equal  (34.  1.)  to  each  of 
the  opposite  sides  DH,  EK :  wherefore  DH  is  equal  to  EK :  add,  or  take 
away  the  common  part  HE  ;  then  DE  is  equal  to  HK :  Wherefore  also 
the  triangle  CDE  is  equal  (38.  1.)  to  the  triangle  BHK :  and  the  parallel- 
ogram DG  is  equal  (36.  1.)  to  the  parallelogram  HN.  For  the  same  rea- 
son, the  triangle  AFG  is  equal  to  the  triangle  LMN,  and  the  parallelogram 
CF  is  equal  (2.  3.  Sup.)  to  the  parallelogram  BM,  and  CG  to  BN;  for 
they  are  opposite.  Therefore  the  planes  which  contain  the  prism  DAG 
are  similar  and  equal  to  those  which  contain  the  prism  HLN,  each  to  each 


200 


SUPPLEMENT  TO  THE  ELEMENTS 


and  the  contiguous  planes  are  also  equally  inclined  to  one  another  (15.  2. 
Sup.),  because  that  the  parallel  planes  AD  and  LH,  as  also  AE  and  LK 


OT 


D 

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are  cut  by  the  same  piane  DN  :  therefore  the  prisms  DAG,  HLN  are 
equal  (1.  3.  Sup.).  If  therefore  the  prism  LNH  be  taken  from  the  solid, 
of  which  the  base  is  the  parallelogram  AB,  and  FDKN  the  plane  opposite 
to  the  base  ;  and  if  from  this  same  solid  there  be  taken  the  prism  AGD; 
the  remaining  solid,  viz.  the  parallelopiped  AH  is  equal  to  the  remaining; 
parallelopiped  AK. 

PROP.  VL     THEOR. 

Solid  parallelopipeds  upon  the  same  base,  and  of  tne  same  altitude,  the  tn^ 
sis  ting  straight  lines  of  which  are  not  terminated  in  the  same  straight  lines 
in  the  plane  opposite  to  the  base,  are  equal  to  one  another. 

Let  the  parallelopipeds  CM,  ON,  be  upon  the  same  base  AB,  and  of  the 
same  altitude,  but  their  insisting  straight  lines  AF,  AG,  LM,  LN,  CD, 
CE,  BH,  BK,  not  terminated  in  the  same  straight  lines  ;  the  solids  CM, 
ON  are  equal  to  one  another. 

Produce  FD,  MH,  and  NG,  KE,  and  let  them  meet  one  another  in  the 
poinds  0,  P,  Q,  R  ;  and  join  AO,  LP,  BQ,  CR.  Because  the  planes  (def. 
5.  3.  Sup.),  LBHM  and  ACDF  are  parallel,  and  because  the  plane  LBHM 
is  that  in  which  are  the  parallels  LB,  MHPQ  (def.  5.  3.  Sup.),  and  in  which 


OF  GEOMETRY.     BOOK  III. 


201 


also  is  the  figure  BLPQ ;  and  because  the  plane  ACDF  is  that  in  which 
are  the  parallels  AC,  FDOR,  and  in  which  also  is  the  figure  CAOR ; 
therefore  the  figures  BLPQ,  CAOR,  are  in  parallel  planes.  In  like  man- 
ner, because  the  planes  ALNG  and  CBKE  are  parallel,  and  the  plane 
ALNG  is  that  in  which  are  the  parallels  AL,  OPGN,  and  in  which  also 
is  the  figure  ALPO  ;  and  the  plane  CBKE  is  that  in  which  are  the  paral- 
lels CB,  RQEK,  and  in  which  also  is  the  figure  CBQR  ;  therefore  the 
figures  ALPO,  CBQR,  are  in  parallel  planes.  But  the  planes  ACBL, 
ORQPare  also  parallel ;  therefore  the  solid  CP  is  a  parallelopiped.  Now 
the  solid  parallelopiped  CM  is  equal  (5.  2.  Sup.)  to  the  solid  parallelopiped 
CP,  because  they  are  upon  the  same  base,  and  their  insisting  straight  lines 
AF,  AO,  CD,  CR  ;  LM,  LP,  BH,  BQ  are  terminated  in  the  same  straight 
lines  FR,  MP ;  and  the  solid  CP  is  equal  (5.  2.  Sup.)  to  the  solid  CN ; 
for  they  are  upon  the  same  base  ACBL,  and  their  insisting  straight  lines 
AO,  AG,  LP,  LN  ;  CR,  CE,  BQ,  BK  are  terminated  in  the  same  straight 
lines  ON,  RK  ;  Therefore  the  solid  CM  is  equal  to  the  solid  CN. 

PROP.  VII.     THEOR. 

Solid  parallelopipedSf  which  are  upon  equal  bases,  and  of  the  same  altitude, 
are  equal  to  one  another. 

Let  the  solid  parallelopipeds,  AE,  CF,  be  upon  equal  bases  AB,  CD, 
and  be  of  the  same  altitude  ;  the  solid  AE  is  equal  to  the  solid  CF. 

Case  1.  Let  the  insisting  straight  lines  be  at  right  angles  to  the  bases 
AB,  CD,  and  let  the  bases  be  placed  in  the  same  plane,  and  so  as  that  the 
sides  CL,  LB,  be  in  a  straight  line ;  therefore  the  straight  line  LM,  which 
is  at  right  angles  to  the  plane  in  which  the  bases  are,  in  the  point  L,  is 
common  (11.  2.  Sup.)  to  the  two  solids  AE,  CF  ;  let  the  other  insisting 
lines  of  the  solids  be  AG,  HK,  BE  ;  DF,  OP,  CN  :  and  first,  let  the  angle 
ALB  be  equal  to  the  angle  CLD  ;  then  AL,  LD  are  in  a  straightline(14. 
1 .).  Produce  OD,  HB,  and  let  them  meet  in  Q  and  complete  the  solid 
parallelopiped  LR,  the  base  of  which  is  the  parallelogram  LQ,  and  of 
which  LM  is  one  of  its  insisting  straight  lines  :  therefore,  because  the  pa- 
rallelogram AB  is  equal  to  CD,  as  the  base  AB  is  to  the  base  LQ,  so  is 
(7.  5.)  the  base  CD  to  the  same  LQ  :  and  because  the  solid  parallelopiped 
AR  is  cut  by  the  plane  LMEB,  which  is  parallel  to  the  opposite  planes 
AK,  DR ;  as  the  base  AB  is  to  the  base  LQ,  so  is  (3.  3.  Sup.)  the  solid 


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26 


202  SUPPLEMENT  TO  THE  ELEMENTS 

AE  to  the  solid  LR  :  for  the  same  reason  because  the  solid  parallelopiped 
CR  is  cut  by  the  plane  LMFD,  which  is  parallel  to  the  opposite  planes 
CP,  BR  ;  as  the  base  CD  to  the  base  LQ  ;  so  is  the  solid  CF  to  the  solid 
LR  ;  but  as  the  base  AB  to  the  base  LQ,  so  the  base  CD  to  the  base  LQ, 
as  has  been  proved :  therefore  as  the  solid  AE  to  .the  solid  LR,  so  is  the 
solid  CF  to  the  solid  LR  ;  and  therefore  the  solid  AE  is  equal  (9.  5.)  to 
the  solid  CF. 

But  let  the  solid  parallelepipeds,  SE,  CF  be  upon  equal  bases  SB,  CD, 
and  be  of  the  same  altitude,  and  let  their  insisting  straight  lines  be  at  right 
angles  to  the  bases ;  and  place  the  bases  SB,  CD  in  the.same  plane,  so 
that  CL,  LB  be  in  a  straight  line  ;  and  let  the  angles  SLB,  CLE),  be  un- 
equal ;  the  solid  SE  is  also  in  this  case  equal  to  the  solid  CF.  Produce 
DL,  TS  until  they  meet  in  A,  and  from  B  draw  BH  parallel  to  DA  ;  and 
let  HB,  CD  produced  meet  in  Q,  and  complete  the  solids  AE,  LR  :  there- 
fore the  solid  AE,  of  which  the  base  is  the  parallelogram  LE,  and  AK  the 
plane  opposite  to  it,  is  equal  (5.  3.  Sup.)  to  the  solid  SE,  of  which  the  base 
is  LE,  and  SX  the  plane  opposite  ;  for  they  are  upon  the  same  base  LE, 
and  of  the  same  altitude,  and  their  insisting  straight  lines,  viz.  LA,  LS, 
BH,  BT  ;  MG,  MU,  EK,  EX,  are  in  the  same  straight  lines  AT,  GX : 
and  because  the  parallelogram  AB  is  equal  (35.  1.)  to  SB,  for  they  are 
upon  the  same  base  LB,  and  between  the  same  parallels  LB,  AT  ;  and 
because  the  base  SB  is  equal  to  the  base  CD  ;  therefore  the  base  AB  is 
equal  to  the  base  CD :  but  the  angle  ALB  is  equal  to  the  angle  CLD : 
therefore,  by  the  first  case,  the  solid  AE  is  equal  to  the  solid  CF  ;  but  the 
solid  AE  is  equal  to  the  solid  SE,  as  was  demonstrated :  therefore  the 
solifl  SE  is  equal  to  the  solid  CF. 

Case  2.     If  the  insisting  straight  lines  AG,  HK,  BE,  LM  ;  CN,  RS, 


DF,  OP,  be  not  at  right  angles  to  the  bases  AB,  CD  ;  in  this  case  likewise 
the  solid  AE  is  equal  to  the  solid  CF.  Because  solid  parallelepipeds  on 
the  same  base,  and  of  the  same  altitude,  are  equal  (6.  3.  Sup.),  if  two  solid 
parallelepipeds  be  constituted  on  the  bases  AB  and  CD  of  the  same  alti- 
tude with  the  solids  AE  and  CF,  and  with  their  insisting  lines  perpendicu- 
lar to  their  bases,  they  will  be  equal  to  the  solids  AE  and  CF  ;  and,  by  the 
first  case  of  this  proposition,  ihey  will  be  equal  to  one  another ;  wherefore, 
the  solids  AE  and  CF  are  also  equal. 


OF  GEOMETRY.    BOOK  III. 


203 


PROP.  VIII.    THEOR. 

Solid  parallelopipeds  which  have  the  same  altitude ^  are  to  one  another  as  their 

bases. 

Let  AB,  CD  be  solid  parallelopipeds  of  the  same  altitude  ;  they  are  to 
one  another  as  their  bases  ;  that  is,  as  the  base  AE  to  the  base  CF,  so  is 
the  solid  AB  to  the  solid  CD. 

To  the  straight  line  FG  apply  the  parallelogram  FH  equal  (Cor.  Prop. 
45.  l.)to  AE,  so  that  the  angle  FGH  be  equal  to  the  angle  LCG;  and 


\/ 

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complete  the  solid  parallelopiped  GK  upon  the  base  FH,  one  of  whose  in- 
sisting lines  is  FD,  whereby  the  solids  CD,  GK  must  be  of  the  same  alti- 
tude. Therefore  the  solid  AB  is  equal  (7.  3.  Sup.)  to  the  solid  GK,  be- 
cause they  are  upon  equal  bases  AE,  FH,  and  are  of  the  same  altitude  : 
and  because  the  solid  parallelopiped  CK  is  cut  by  the  plane  DG  which  is 
parallel  to  its  opposite  planes,  the  base  HF  is  (3.  3.  Sup.)  to  the  base  FC, 
as  the  solid  HD  to  the  solid  DC  :  But  the  base  HF  is  equal  to  the  base 
AE,  and  the  solid  GK  to  the  solid  AB  :  therefore,  as  the  base  AE  to  the 
base  CF,  so  is  the  solid  AB  to  the  solid  CD. 

CoR.  1 .  From  this  it  is  manifest,  that  prisms  upon  triangular  bases,  and 
of  the  same  altitude,  are  to  one  another  as  their  bases.  Let  the  prisms 
BNM,  DPG,  the  bases  of  which  are  the  triangles  AEM,  CFG,  have  the 
same  altitude  :  complete  the  parallelograms  AE,  CF,  and  the  solid  paral 
lelopipeds  AB,  CD,  in  the  first  of  which  let  AN,  and  in  the  other  let  CP 
be  one  of  the  insisting  lines.  And  because  the  solid  parallelopipeds  AB, 
CP  have  the  same  altitude,  they  are  to  one  another  as  the  base  AE  is  to 
the  base  CF  ;  wherefore  the  prisms,  which  are  their  halves  (4.  3.  Sup.) 
are  to  one  another,  as  the  base  AE  to  the  base  CF  ;  that  is,  as  the  trian- 
gle AEM  to  the  triangle  CFG. 

CoR.  2.  Also  a  prism  and  a  parallelopiped,  which  have  the  same  alti- 
tude, are  to  one  another  as  their  bases  ;  that  is,  the  prism  BNM  is  to  the 
parallelopiped  CD  as  the  triangle  AEM  to  the  parallelogram  LG.  For 
by  the  last  Cor.  the  prism  BNM  is  to  the  prism  DPG  as  the  triangle  AME 
to  the  triangle  CGF,  and  therefore  the  prism  BNM  is  to  twice  the  prism 
DPG  as  the  triangle  AME  to  twice  the  triangle  CGF  (4.  5.) ;  that  is,  the 
prism  BNM  is  to  the  parallelopiped  CD  as  the  triangle  AME  to  the  paral- 
lelogram LG. 


204 


SUPPLEMENT  TO  THE  ELEMENTS 


PROP.  IX.    THEOR. 

Solid  parallelopipeds  are  to  one  another  in  the  ratio  that  is  compounded  of  the 
ratios  of  the  areas  of  their  bases,  and  of  their  altitudes. 

Let  AF  and  GO  be  two  solid  parallelopipeds,  of  which  the  bases  are  the 
parallelograms  AC  and  GK,  and  the  altitudes,  the  perpendiculars  let  fall 
on  the  planes  of  these  bases  from  any  point  in  the  opposite  planes  EF  and 
MO  ;  the  solid  AF  is  to  the  solid  GO  in  a  ratio  compounded  of  the  ratios 
of  the  base  AC  to  the  base  GK,  and  of  the  perpendicular  on  AC,  to  the 
perpendicular  on  GK. 

Case  1.  When  the  insisting  lines  are  perpendicular  to  the  bases  AC 
and  GK,  or  when  the  solids  are  upright. 

In  GM,  one  of  the  insisting  lines  of  the  solid  GO,  take  GQ  equal  to  AE, 
one  of  the  insisting  lines  of  the  solid  AF,  and  through  Q  let  a  plane  pass 
parallel  to  the  plane  GK,  meeting  the  other  insisting  lines  of  the  solid  GO 


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in  the  points  R,  S  and  T.  It  is  evident  that  GS  is  a  solid  parallelepiped 
(def.  5.  3.  Sup.)  and  that  it  has  the  same  altitude  with  AF,  viz.  GQ  or 
AE.  Now  the  solid  AF  is  to  the  solid  GO  in  a  ratio  compoimded  of  the 
ratios  of  the  solid  AF  to  the  solid  GS  (def.  10.  5.),  and  of  the  solid  GS  to 
the  solid  GO ;  but  the  ratio  of  the  solid  AF  to  the  solid  GS,  is  the  same 
"with  that  of  the  base  AC  to  the  base  GK  (8.  3.  Sup.),  because  their  alti- 
tudes AE  and  GQ  are  equal ;  and  the  ratio  of  the  solid  GS  to  the  solid 
GO,  is  the  same  with  that  of  GQ  to  GM  (3.  2.  Sup.) ;  therefore,  the  ratio 
which  is  compounded  of  the  ratios  of  the  solid  AF  to  the  solid  GS,  and  of 
the  solid  GSto  the  solid  GO,  is  the  same  with  the  ratio  which  is  compound- 
ed of  the  ratios  of  the  base  AC  to  the  base  GK,  and  of  the  altitude  AE  to 
the  altitude  GM  (F.  5.).  But  the  ratio  of  the  solid  AF  to  the  solid  GO,  is 
that  which  is  compounded  of  the  ratios  of  AF  to  GS,  and  of  GS  to  GO ; 
therefore,  the  ratio  of  the  solid  AF  to  the  solid  GO  is  compounded  of  the 
ratios  of  the  base  AC  to  the  base  GK,  and  of  the  altitude  AE  to  the  alti- 
tude GM. 

Case  2.     When  the  insisting  lines  are  not  perpendicular  to  the  bases. 


OF  GEOMETRY.    BOOK  III.  205 

Let  the  parallelograms  AC  and  GK  be  the  bases  as  before,  and  let  AE 
and  GM  be  the  altitudes  of  two  parallelopipeds  Y  and  Z  on  these  bases. 
Then,  if  the  upright  parallelopipeds  AF  and  GO  be  constituted  on  the 
bases  AC  and  GK,  with  the  altitudes  AE  and  GM,  they  will  be  equal  to 
the  parallelopipeds  Y  and  Z  (7.  3.  Sup.).  Now,  the  solids  AF  and  GO, 
by  the  first  case,  are  in  the  ratio  compounded  of  the  ratios  of  the  bases  AC 
and  GK,  and  of  the  altitudes  AE  and  GM  ;  therefore  also  the  solids  Y 
and  Z  have  to  one  another  a  ratio  that  is  compounded  of  the  same  ratios. 

Cor.  1.  Hence,  two  straight  lines  may  be  found  having  the  same  ratio 
with  the  two  parallelopipeds  AF  and  GO.  To  AB,  one  of  the  sides  of  the 
parallelogram  AC,  apply  the  parallelogram  BV  equal  to  GK,  having  an 
angle  equal  to  the  angle  BAD  (Prop.  44.  1.) ;  and  as  AE  to  GM,  so  let 
A  V  be  to  AX  (12.  6.),  then  AD  is  to  AX  as  the  solid  AF  to  the  solid  GO. 
For  the  ratio  of  AD  to  AX  is  compounded  of  the  ratios  (def.  10.  5.)  of  AD 
to  AV,  and  of  AV  to  AX  ;  but  the  ratio  of  AD  to  AV  is  the  same  with 
that  of  the  parallelogram  AC  to  the  parallelogram  BV  (1.  6.)  or  GK ; 
and  the  ratio  of  AY  to  AX  is  the  same  with  that  of  AE  to  GM ;  therefore 
the  ratio  of  AD  to  AX  is  compounded  of  the  ratios  of  AC  to  GK,  and  of 
AE  to  GM  (E.  5.).  But  the  ratio  of  the  solid  AF  to  the  solid  GO  is  com- 
pounded of  the  same  ratios ;  therefore,  as  AD  to  AX,  so  is  the  solid  AF  to 
the  solid  GO. 

CoR.  2.  If  AF  and  GO  are  two  parallelopipeds,  and  if  to  AB,  to  the 
perpendicular  from  A  upon  DC,  and  to  the  altitude  of  the  parallelepiped 
AF,  the  numbers  L,  M,  N,  be  proportional :  and  if  to  AB,  to  GH,  to  the 
perpendicular  from  G  on  LK,  and  to  the  altitude  of  the  parallelepiped  GO, 
the  numbers  L,  Z,  m,  n,  be  proportional ;  the  solid  AF  is  to  the  solid  GO 
as  LxMxN  to  IXmXn. 

For  it  may  be  proved,  as  in  the  7th  of  the  1st  of  the  Sup.  that  L  X  Mx 
N  is  to  Z  X  m  X  w  in  the  ratio  compounded  of  the  ratio  of  L  X  M  to  /  x  ?w,  and 
of  the  ratio  of  N  to  n.  Now  the  ratio  of  L  X  M  to  Zx  ?w  is  that  of  the  area 
of  the  parallelogram  AC  to  that  of  the  parallelogram  GK  ;  and  the  ratio 
of  N  to  n  is  the  ratio  of  the  altitudes  of  the  parallelopipeds,  by  hypothesis, 
therefore,  the  ratio  ofLxMxNto  IXmXn  is  compounded  of  the  ratio  of 
the  areas  of  the  bases,  and  of  the  ratio  of  the  altitudes  of  the  parallelopipeds 
AF  and  GO  ;  and  the  ratio  of  the  parallelopipeds  themselves  is  shewn,  in 
this  proposition,  to  be  compounded  of  the  same  ratios  ;  therefore  it  is  the 
same  with  that  of  the  product  L  X  M X  N  to  the  product  IxmXn. 

Cor.  3.  Hence  all  prisms  are  to  one  another  in  the  ratio  compounded 
of  the  ratios  of  their  bases,  and  of  their  altitudes.  For  every  prism  is 
equal  to  a  parallelepiped  of  the  same  altitude  with  it,  and  of  an  equal  base 
(2.  Cor.  8.  3.  Sup.). 

PROP.  X.     THEOR. 

Solid  parallelopipeds,  which  have  their  bases  and  altitudes  reciprocally  propor- 
tional, are  equal ;  and  parallelopipeds  which  are  equal,  have  their  bases  and 
altitudes  reciprocally  proportional. 

Let  AG  and  KQ  be  two  solid  parallelopipeds,  of  which  the  bases  are 


206 


SUPPLEMENT  TO  THE  ELEMENTS 


AC  and  KM,  and  the  altitudes  AE  and  KO,  and  let  AC  be  to  KM  as  KO 
to  AE  ;  the  solids  AG  and  KQ  are  equal. 

As  the  base  AC  to  the  base  KM,  so  let  the  straight  line  KO  be  to  the 
straight  line  S.  Then,  since  AC  is  to  KM  as  KO  to  S,  and  also  by  hypo- 
thesis, AC  to  KM  as  KO  to  AE,  KO  has  the  same  ratio  to  S  that  it  has 
to  AE  (11.  5.)  ;  wherefore  AF  is  equal  to  S  (9.  5.).     But  the  solid  AG  is 


to  the  solid  KQ,  in  the  ratio  compounded  of  the  ratios  of  AE  to  KO,  and 
of  AC  to  KM  (9.  3.  Sup.),  that  is,  in  the  ratio  compounded  of  the  ratios  of 
AE  to  KO,  and  of  KO  to  S.  And  the  ratio  of  AE  to  S  is  also  compound- 
ed of  the  same  ratios  (def.  10.  5.) ;  therefore,  the  solid  AG  has  to  the  solid 
KQ  the  same  ratio  that  AE  has  to  S.  But  AE  was  proved  to  be  equal  to 
S,  therefore  AG  is  equal  to  KQ. 

Again,  if  the  solids  AG  and  KQ  be  equal,  the  base  AC  is  to  the  base 
KM  as  the  altitude  KO  to  the  altitude  AE.  Take  S,  so  that  AC  may  be 
to  KM  as  KO  to  S,  audit  will  be  shewn,  as  was  done  above,  that  the  solid 
AG  is  to  the  solid  KQ  as  AE  to  S  ;  now,  the  solid  AG  is,  by  hypothesis, 
equal  to  the  solid  KQ  :  therefore,  AE  is  equal  to  S  ;  but,  by  construction, 
AC  is  to  KM,  as  KO  is  to  S  ;  therefore,  AC  is  to  KM  as  KO  to  AE. 

CoR.  In  the  same  manner,  it  may  be  demonstrated,  that  equal  prisms 
have  their  bases  and  altitudes  reciprocally  proportional,  and  conversely. 

PROP.  XL     THEOR. 

Similar  solid  parallelopipeds  are  to  one  another  in  the  triplicate  ratio  of  their 

homologous  sides. 

Let  AG,  KQ  be  two  similar  parallelopipeds,  of  which  AB  and  KL  are 
two  homologous  sides  ;  the  ratio  of  the  solid  AG  to  the  solid  KQ  is  tripli- 
cate of  the  ratio  of  AB  to  KL. 

Because  the  solids  are  similar,  the  parallelograms  AF,  KP  are  similar 
(def.  2.  3.  Sup.),  as  also  the  parallelograms  AH,  KR  ;  therefore,  the  ratios 
of  AB  to  KL,  of  AE  to  KO,  and  of  AD  to  KN  are  all  equal  (def.  1.  6.). 
But  the  ratio  of  the  solid  AG  to  the  solid  KQ  is  compounded  of  the  ratios 
of  AC  to  KM,  and  of  AE  to  KO.  Now,  the  ratio  of  AC  to  KM,  because 
they  are  equiangular  parallelograms,  is  compounded  (23.  6.)  of  the  ratios 
of  AB  to  KL,  and  of  AD  to  KN.     Wherefore,  the  ratio  of  AG  to  KQ  is 


OF  GEOMETRY.     BOOK  III. 


207 


n 

[ 

^ 

\ 

E 

\p 

-TX 

r, 

JJ 

\ 

\ 

Ji 

L. 

B 

isr 


a 


.m: 


compounded  of  the  three  ratios  of  AB  to  KL,  AD  to  KN,  and  AE  to  KO  : 
and  the  three  ratios  have  already  been  proved  to  be  equal ;  therefore,  the 
ratio  that  is  compounded  of  them,  viz.  the  ratio  of  the  solid  AG  to  the  solid 
KQ,  is  triplicate  of  any  of  them  (def.  12.  5.) :  it  is  therefore  triplicate  of 
the  ratio  of  AB  to  KL. 

Cor.  1 .  If  as  AB  to  KL,  so  KL  to  m,  and  as  KL  to  m,  so  is  m  to  n,  then 
AB  is  to  n  as  the  solid  AG  to  the  solid  KQ.  For  the  ratio  of  AB  to  n  is 
triplicate  of  the  ratio  of  AB  to  KL  (def.  12.  5.),  and  is  therefore  equal  to 
that  of  the  solid  AG  to  the  solid  KQ. 

Cor.  2.  As  cubes  are  similar  solids,  therefore  the  cube  on  AB  is  to  the 
cube  on  KL  in  the  triplicate  ratio  of  AB  to  KL,  that  is  in  the  same  ratio 
with  the  solid  AG,  to  the  solid  KQ.  Similar  solid  parallelopipeds  are  there- 
fore to  one  another  as  the  cubes  on  their  homologous  sides. 

Cor.  3.  In  the  same  manner  it  is  proved,  that  similar  prisms  are  to  one 
another  in  the  triplicate  ratio,  or  in  the  ratio  of  the  cubes  of  their  homolo- 
gous sides. 


PROP.  XII.    THEOR. 


If  two  triangular  pyramids  1  which  have  equal  bases  and  altitudes,  be  cut  by  planes 
that  are  parallel  to  the  bases,  and  at  equal  distances  from  them,  the  sections 
are  equal  to  one  another. 

Let  ABCD  and  EFGH  be  two  pyramids,  having  equal  bases  BDC  and 
FGH,  and  equal  altitudes,  viz.  the  perpendiculars  AQ,  and  ES  drawn  from 
A  and  E  upon  the  planes  BDC  and  FGH  :  and  let  them  be  cut  by  planes 
parallel  to  BDC  and  FGH,  and  at  equal  altitudes  QR  and  ST  above  those 
planes,  and  let  the  sections  be  the  triangles  KLM,  NOP  ;  KLM  and  NOP 
are  equal  to  one  another. 

Because  the  plane  ABD  cuts  the  parallel  planes  BDC,  KLM,  the  com- 
mon sections  BD  and  KM  are  parallel  (14.  2.  Sup.).  For  the  same  rea- 
son, DC  and  ML  are  parallel.  Since  therefore  KM  and  ML  are  parallel 
to  BD  and  DC,  each  to  each,  though  not  in  the  same  plane  with  them,  the 
angle  KLM  is  equal  to  the  angle  BDC  (9.  2.  Sup.).  In  like  manner  the 
other  angles  of  these  triangles  are  proved  to  be  equal ;  therefore,  the  trian- 
gles are  equiangular,  and  consequently  similar  ;  and  the  same  is  true  of  the 
triangles  NOP,  FGH. 

Now,  since  the  straight  lines  ARQ,  AKB  meet  the  parallel  planes  BDC 


208 


SUPPLEMENT  TO  THE  ELEMENTS 


and  KML,  they  are  cut  by  them  proportionally  (16. 2.  Sup.),  or  QR  :  RA 
:  :  BK  :  KA ;  "^and  AQ  :  AR  : :  AB  :  AK  (18.  5.),  for  the  same  reason, 
ES  :  ET  :  :  EF  :  EN ;  therefore  AB  :  AK  : :  EF  :  EN,  because  AQ  is 
equal  to  ES,  and  AR  to  ET.  Again,  because  the  triangles  ABC,  AKL 
are  similar, 

AB  :  AK  :  :  BC  :  KL  ;  and  for  the  same  reason 

EF  :  EN  :  :  FG  :  NO;  therefore, 

BC  :  KL  :  :  FG  :  NO.  And,  when  four  straight  lines  are  propor- 
tionals, the  similar  figures  described  on  them  are  proportionals  (22.  6.) ; 
therefore  the  triangle  BCD  is  to  the  triangle  KLM  as  the  triangle  FGH 
to  the  triangle  NOP  ;  but  the  triangle  BDC,  FGH  are  equal ;  therefore, 
the  triangle  KLM  is  also  equal  to  the  triangle  NOP  (1.  5.). 

CoR.  1.  Because  it  has  been  shewn  that  the  triangle  KLM  is  similar 
to  the  base  BCD  ;  therefore,  any  section  of  a  triangular  p}Tamid  parallel 
to  the  base,  is  a  triangle  similar  to  the  base.  And  in  the  same  manner  it  is 
shewn,  that  the  sections  parallel  to  the  base  of  a  polygonal  pyramid  are 
similar  to  the  base. 

CoR.  2.  Hence  also,  in  polygonal  pyramids  of  equal  bases  and  altitudes, 
the  sections  parallel  to  the  bases,  and  at  equal  distances  from  them,  are 
equal  to  one  another. 


PROP.  XIH.    THEOR. 

A  scries  of  prisms  of  the  same  altitude  mai/  be  circ7imscribed  about  any  pyramid, 
such  that  the  sum  of  the  prisms  shall  exceed  the  pyramid  by  a  solid  less  than 
any  given  solid. 

Let  ABCD  be  a  pjnramid,  and  Z*  a  given  solid ;  a  series  of  prisms  hav- 
ing all  the  same  altitude,  may  be  circumscribed  about  the  pyramid  ABCD, 
so  that  their  sum  shall  exceed  ABCD,  by  a  solid  less  than  Z. 


The  solid  Z  ii  not  represented  in  the  figure  of  this,  or  the  following  Proposition. 


OF  GEOMETRY.    BOOK  III. 


209 


Let  Z  be  equal  to  a  prism  standing  on  tlie  same  base  with  the  pyramid, 
viz.  the  triangle  BCD,  and  having  for  its  altitude  the  perpendicular  drawn 
from  a  certain  point  E  in  the  line  AC 
npon  the  plane  BCD.  It  is  evident,  that 
CE  multiplied  by  a  certain  number  m 
will  be  greater  than  AC  ;  divide  CA  into 
as  many  equal  parts  as  there  are  units  in 
m,  and  let  these  be  CF,  FG,  GH,  HA, 
each  of  which  will  be  less  than  CE. 
Through  each  of  the  points  F,  G,  H,  let 
planes  be  made  to  pass  parallel  to  the 
plane  BCD,  making  with  the  sides  of  the 
pyramid  the  sections  FPQ,  GRS,  HTU, 
which  will  be  all  similar  to  one  another, 
and  to  the  base  BCD  (1.  cor.  12.3.  Sup.). 
From  the  point  B  draw  in  the  plane  of 
the  triangle  ABC,  the  straight  line  BK 
parallel  to  CF  meeting  FP  produced  in 
K.  In  like  manner,  from  D  draw  DL  pa- 
rallel to  CF,  meeting  FQ  in  L  :  Join  KL, 
and  it  is  plain,  that  the  solid  KBCDLF 
is  a  prism  (def.  4.  3.  Sup.).  By  the  same 
construction,  let  the  prisms  PM,  RO,  TV 
be  described.  Also,  let  the  straight  line  IP,  which  is  in  the  plane  of  the 
triangle  ABC,  be  produced  till  it  meet  BC  in  h  ;  and  let  the  line  MQ  be 
produced  till  it  meet  DC  in  g :  Join  hg ;  then  hC  gQFP  is  a  prism,  and  is 
equal  to  the  prism  PM  (1.  Cor.  8.  3.  Sup.).  In  the  same  manner  is  describ- 
ed the  prism  mS  equal  to  the  prism  RO,  and  the  prism  qlJ  equal  to  the 
prism  TV.  The  sum,  therefore,  of  all  the  inscribed  prisms  hQ,  mS,  and 
qU  is  equal  to  the  sum  of  the  prisms  PM,  RO  and  TV,  that  is,  to  the  sum 
of  all  the  circumscribed  prisms  except  the  prism  BL ;  wherefore,  BL  is  the 
excess  of  the  prism  circumscribed  about  the  pyramid  ABCD  above  the 
prisms  inscribed  within  it.  But  the  prism  BL  is  less  than  the  prism  which 
has  the  triangle  BCD  for  its  base,  and  for  its  altitude  the  perpendicular 
from  E  upon  the  plane  BCD  ;  and  the  prism  which  has  BCD  for  \\s  base, 
and  the  perpendicular  from  E  for  its  altitude,  is  by  hypothesis  equal  to  the 
given  solid  Z  ;  therefore  the  excess  of  the  circumscribed,  above  the  inscrib- 
ed prisms,  is  less  than  the  given  solid  Z.  But  the  excess  of  the  circum- 
scribed prisms  above  the  inscribed  is  greater  than  their  excess  above  the 
pyramid  ABCD,  because  ABCD  is  greater  than  the  sum  of  the  inscribed 
prisms.  Much  more,  therefore,  is  the  excess  of  the  circumscribed  prisms 
above  the  pyramid,  less  than  the  solid  Z.  A  series  of  prisms  of  the  same 
altitude  has  therefore  been  circumscribed  about  the  pyramid  ABCD,  ex- 
ceeding it  by  a  solid  less  than  the  given  solid  Z. 


PROP.  XIV.    THEOR 


Pyramids  that  have  equal  bases  and  altitudes  are  equal  to  one  another. 

Let  ABCD,  EFGH,  be  two  pyramids  that  have  equal  bases  BCD,  FGH 

27 


210 


SUPPLEMENT  TO  THE  ELEMENTS 


and  also  equal  altitudes,  viz.  the  perpendiculars  drawn  from  the  vertices  A 
and  E  upon  the  planes  BCD,  FGH  :  the  pyramid  ABCD  is  equal  to  the 
pyramid  EFGH. 

If  they  are  not  equal,  let  the  pjTamid  EFGH  exceed  the  pyramid  ABCD 
by  the  solid  Z.  Then,  a  series  of  prisms  of  the  same  altitude  may  be  de 
scribed  about  the  pyramid  ABCD  that  shall  exceed  it,  by  a  solid  less  than 
Z  (13.  3.  Sup.) ;  let  these  be  the  prisms  that  have  for  their  bases  the  trian- 
gles BCD,  NQL,  ORI,  PSM.  Divide  EH  into  the  same  number  of  equal 
parts  into  which  AD  is  divided,  viz.  HT,  TU,  UV,  YE,  and  through  the 


points  T,  U  and  V,let  the  sections  TZW,  U^X,  V(?^Ybe  made  parallel 
to  the  base  FGH.  The  section  NQL  is  equal  to  the  section  WZT  (12. 
3.  Sup.) ;  as  also  ORI  to  X^U,  and  PSM  to  Y<Z'V ;  and  therefore  also  the 
prisms  that  stand  upon  the  equal  sections  are  equal  (1.  Cor.  8.  3.  Sup.), 
that  is,  the  prism  which  stands  on  the  base  BCD,  and  which  is  between 
the  planes  BCD  and  NQL,  is  equal  to  the  prism  which  stands  on  the  base 
FGH,  and  which  is  between  the  planes  FGH  and  WZT  ;  and  so  of  the 
rest,  because  they  have  the  same  altitude  :  wherefore,  the  sum  of  all  the 
prisms  described  about  the  pyramid  ABCD  is  equal  to  the  sum  of  all  those 
described  about  the  pyramid  EFGH.  But  the  excess  of  the  prisms  de- 
scribed about  the  pyramid  ABCD  above  the  pyramid  ABCD  is  less  than 
Z(13.  3.  Sup.);  and  therefore,  the  excess  of  the  prism  described  about 
the  pyramid  EFGH  above  the  pyramid  ABCD  is  also  less  than  Z.  But 
the  excess  of  the  pyramid  EFGH  above  the  pyramid  ABCD  is  equal  to 
Z,  by  hypothesis,  therefore,  the  pyramid  EFGH  exceeds  the  pyramid 
ABCD,  more  than  the  prisms  described  about  EFGH  exceeds  the  same 
pyramid  ABCD.  The  pyramid  EFGH  is  therefore  greater  than  the  sum 
of  the  prisms  described  about  it,  which  is  impossible.  The  pyramids 
ABCD,  EFGBL  therefore,  are  not  unequal,  that  is,  they  are  equal  to  one 
another. 


OF  GEOMETRY.    BOOK  III. 


211 


PROP.  XV.     THEOR. 

Every  prism  having  a  triangular  base  may  be  divided  into  tnree  pyramids  that 
have  triangular  bases,  and  that  are  equal  to  another. 

Let  there  be  a  prism  of  which  the  base  is  the  triangle  ABC,  and  let 
DEF  be  the  triangle  opposite  the  base  :  The  prism  ABGDEF  may  be 
divided  into  three  equal  pyramids  having  triangular  bases. 

Join  AE,  EC,  CD  ;  and  because  ABED  is  a  parallelogram,  of  which 
AE  is  the  diameter,  the  triangle  ADE  is  equal 
(34.  1.)  to  the  triangle  ABE  :  therefore  the  py- 
ramid of  which  the  base  is  the  triangle  ADE, 
and  vertex  the  point  C,is  equal  (14.  3.  Sup.)  to 
the  pyramid,  of  which  the  base  is  the  triangle 
ABE,  and  vertex  the  point  C.  But  the  pyra- 
mid of  which  the  base  is  the  triangle  ABE,  and 
vertex  the  point  C,  that  is,  the  pyramid  ABCE 
is  equal  to  the  pyramid  DEFC  (14.  3.  Sup.), 
for  they  have  equal  bases,  viz.  the  triangles  • 
ABC,  DEF,  and  the  same  altitude,  viz.  the  al- 
titude of  the  prism  ABCDEF.  Therefore  the 
three  pyramids  ADEC,  ABEC,  DFEC  are 
equal  to  one  another.  But  the  pyramids  ADEC, 
ABEC,  DFEC  make  up  the  whole  prism 
ABCDEF  ;  therefore,  the  prism  ABCDEF  is 
divided  into  three  equal  pyramids. 

CoR.  1.  From  this  it  is  manifest,  that  every  pyramid  is  the  third  part 
of  a  prism  which  has  the  same  base,  and  the  same  altitude  with  it ;  for  if 
the  base  of  the  prism  be  any  other  figure  than  a  triangle,  it  may  be  divided 
into  prisms  having  triangular  bases. 

CoR.  2.  Pyramids  of  equal  altitudes  are  to  one  another  as  their  bases  ; 
because  the  prisms  upon  the  same  bases,  and  of  the  same  altitude,  are  (1. 
Cor.  8. 3.  Sup.)  to  one  another  as  their  bases. 


PROP.  XVI.     THEOR. 

If  from  any  "point  in  the  circumference  of  the  base  of  a  cylinder,  a  straight 
line  be  drawn  perpendicular  to  the  plane  of  the  base,  it  will  be  wholly  in  the 
cylindric  superficies. 

Let  ABCD  be  a  cylinder  of  which  the  base  is  the  circle  AEB,  DFC 
the  circle  opposite  to  the  base,  and  GH  the  axis  ;  from  E,  any  point  in  the 
circumference  AEB,  let  EF  be  drawn  perpendicular  to  the  plane  of  the 
circle  AEB  :  the  straight  line  EF  is  in  the  superficies  of  the  cylinder. 

Let  F  be  the  point  in  which  EF  meets  the  plane  DFC  opposite  to  the 
base ;  join  EG  and  FH  ;  and  let  AGHD  be  the  rectangle  (14.  def.  3, 
Sup.)  by  the  revolution  of  which  the  cylinder  ABCD  is  described. 


212 


SUPPLEMENT  TO  THE  ELEMENTS 


Now,  because  GH  is  at  right  angles  to  GA, 
the  straight  line,  which  by  its  revolution  des- 
cribes the  circle  AEB,  it  is  at  right  angles  to 
all  the  straight  lines  in  the  plane  of  that  circle 
which  meet  it  in  G,  and  it  is  therefore  at  right 
angles  to  the  plane  of  the  circle  AEB.  But 
EF  is  at  right  angles  to  the  same  plane ;  there- 
fore, EF  and  GH  are  parallel  (6.  2.  Sup.)  and 
in  the  same  plane.  And  since  the  plane  through 
GH  and  EF  cuts  the  parallel  planes  AEB, 
DFC,  in  the  straight  lines  EG  and  FH,  EG  is 
parallel  to  FH  (14.  2.  Sup.).  The  figure 
EGHF  is  therefore  a  parallelogram,  and  it  has 
the  angle  EGH  a  right  angle,  therefore  it  is  a 
rectangle,  and  is  equal  to  the  rectangle  AH, 
because  EG  is  equal  to  AG.  Therefore,  when 
in  the  revolution  of  the  rectangle  AH,  the  straight  line  AG  coincides  with 
EG,  the  two  rectangles  AH  and  EH  will  coincide,  and  the  straight  line 
AD  will  coincide  with  the  straight  line  EF.  But  AD  is  always  in  the 
superficies  of  the  cylinder,  for  it  describes  that  superficies  ;  therefore,  EF 
is  also  in  the  superficies  of  the  cylinder. 


PROP.  XVn.    THEOR. 

A  cylinder  and  a  parallelopiped  having  equal  bases  and  altitudes,  are  equal  to 

one  another. 


Let  ABCD  be  a  cylinder,  and  EF  a  parallelopiped  having  equal  bases, 
viz.  the  circle  AGB  and  the  parallelogram  EH,  and  having  also  equal  al- 
titudes ;  the  cylinder  ABCD  is  equal  to  the  parallelopiped  EF. 


QS 


B 


:e:rk 


If  not,  let  them  be  unequal ;  and  first,  let  the  cylinder  be  less  than  the 
parallelopiped  EF  ;  and  from  the  parallelopiped  EF  let  there  be  cut  oflf  a 


OF  GEOMETRY.    BOOK.  III. 


213 


part  EQ  bj^  a  plane  PQ  parallel  to  NF,  equal  to  the  cylinder  ABCD.  In 
the  circle  AGB  inscribe  the  polygon  AGKBLM  that  shall  differ  from  the 
circle  by  a  space  less  than  the  parallelogram  PH  (Cor.  1.4.  1.  Sup.),  and 
cut  off  from  the  parallelogram  EH,  a  part  OR  equal  to  the  polygon 
AGKBLM.  The  point  R  will  fall  between  P  and  N.  On  the  polygon 
AGKBLM  let  an  upright  prism  AGBCD  be  constituted  of  the  same  alti- 
tude with  the  cylinder,  which  will  therefore  be  less  than  the  cylinder,  be- 
cause it  is  within  it  (16.  3.  Sup.) ;  and  if  through  the  point  R  a  plane  RS 
parallel  to  NF  be  made  to  pass,  it  will  cut  off  the  parallelepiped  ES  equal 
(2.  Cor.  8.  3.  Sup.)  to  the  prism  AGBC,  because  its  base  is  equal  to  that 
of  the  prism,  and  its  altitude  is  the  same.  But  the  prism  AGBC  is  less 
than  the  cylinder  ABCD,  and  the  cylinder  ABCD  is  equal  to  the  parallel- 
epiped EQ,  by  hypothesis  ;  therefore,  ES  is  less  than  EQ,  and  it  is  also 
greater,  which  is  impossible.  The  cylinder  ABCD,  therefore,  is  not  less 
than  the  parallelepiped  EF ;  and  in  the  same  manner,  it  may  be  shewn 
not  to  be  greater  than  EF. 

PROP.  XVIII.     THEOR. 


If  a  cone  and  cylinder  have  the  same  base  and  the  same  altitude^  the  cone  is  the 
third  part  of  the  cylinder. 

Let  the  cone  ABCD,  and  the  cylinder  BFKG  have  the  same  base,  viz. 
the  circle  BCD,  and  the  same  altitude,  viz.  the  perpendicular  from  the 
point  A  upon  the  plane  BCD,  the  cone  ABCD  is  the  third  part  of  the  cylin- 
der BFKG. 

If  not,  let  the  cone  ABCD  be  the  third  part  of  another  cylinder  LMNO, 
having  the  same  altitude  with  the  cylinder  BFKG,  but  let  the  bases  BCD 
and  LIM  be  unequal ;  and  first,  let  BCD  be  greater  than  LIM. 


Then,  because  the  circle  BCD  is  greater  than  the  circle  LIM,  a  polygon 
maybe  inscribed  in  BCD,  that' shall  differ  from  it  less  than  LIM  does  (4. 
1.  Sup.),  and  which,  therefore,  will  be  greater  than  LIM.  Let  this  be  the 
polygon  BECFD  ;  and  upon  BECFD,  let  there  be  constituted  the  pyra- 
mid ABECFD,  and  the  prism  BCFKHG. 


214  SUPPLEMENT  TO  THE  ELEMENTS 

Because  the  polygon  BECFD  is  greater  than  the  circle  LIM,  the  prism 
BCFKHG  is  greater  than  the  cylinder  LMNO,  for  they  have  the  same 
altitude,  but  the  prism  has  the  greater  base.  But  the  pyramid  ABECFD 
is  the  third  part  of  the  prism  (15.  3.  Sup.)  BCFKHG,  therefore  it  is  great- 
er than  the  third  part  of  the  cylinder  LMNO.  Now,  the  cone  ABECFD 
is,  by  hypothesis,  the  third  part  of  the  cylinder  LMNO,  therefore  the  pyra- 
mid ABECFD  is  greater  than  the  cone  ABCD,  and  it  is  also  less,  because 
it  is  inscribed  in  the  cone,  which  is  impossible.  Therefore,  the  cone  ABCD 
is  not  less  than  the  third  part  of  the  cylinder  BFKG :  And  in  the  same 
manner,  by  circumscribing  a  polygon  about  the  circle  BCD,  it  may  be 
shewn  that  the  cone  i\.BCD  is  not  greater  than  the  third  part  of  the  cylin- 
der BFKG ;  therefore,  it  is  equal  to  the  third  part  of  that  cylinder, 

PROP.  XIX.    THEOR. 

If  a  hemisphere  and  a  cone  have  equal  bases  and  altitudes,  a  series  of  cylinders 
may  he  inscribed  in  the  hemisphere,  and  another  series  may  be  described  about 
the  cone,  having  all  the  same  altitudes  with  one  another,  and  such  that  their 
sum  shall  differ  from  the  sum  of  the  hemisphere,  and  the  cone,  by  a  solid 
less  than  any  given  solid. 

Let  ADB  be  a  semicircle  of  which  the  centre  is  C,  and  let  CD  be  at  right 
angles  to  AB  ;  let  DB  and  DA  be  squares  described  on  DC,  draw  CE, 
and  let  the  figure  thus  constructed  revolve  about  DC :  then,  the  sector 
BCD,  which  is  the  half  of  the  semicircle  ADB,  will  describe  a  hemisphere 
having  C  for  its  centre  (7  def.  3.  Sup.),  and  the  triangle  CDE  will  describe 
a  cone,  having  its  vertex  to  C,  and  having  for  its  base  the  circle  (11.  def. 
3.  Sup.)  described  by  DE,  equal  to  that  described  by  BC,  which  is  the  base 
of  the  hemisphere.  Let  W  be  any  given  solid.  A  scries  of  cylinders  may 
be  inscribed  in  the  hemisphere  ADB,  and  another  described  about  the  cone 
ECI,  so  that  their  sum  shall  differ  from  the  sum  of  the  hemisphere  and 
the  cone,  by  a  solid  less  than  the  solid  W. 

Upon  the  base  of  the  hemisphere  let  a  cylinder  be  constituted  equal  to 
W,  and  let  its  altitude  be  CX.  Divide  CD  into  such  a  number  of  equal ' 
parts,  that  each  of  them  shall  be  less  than  CX  ;  let  these  be  CH,  HG,  GF, 
and  FD.  Through  the  points  F,  G,  H,  draw  FN,  GO,  HP  parallel  to 
CB,  meeting  the  circle  in  the  points  K,  L  and  M ;  and  the  straight  line 
CE  in  the  points  Q,  R  and  S.  From  the  points  K,  L,  M  draw  Kf,  Lg, 
Mh,  perpendicular  to  GO,  HP  and  CB  ;  and  from  Q,  R,  and  S,  draw  Qq, 
Rr,  Ss,  perpendicular  to  the  same  lines.  It  is  evident,  that  the  figure  being 
thus  constructed,  if  the  whole  revolve  about  CD,  the  rectangles  Ff,  Gg,  Hh 
will  describe  cylinders  (14.  def.  3.  Sup.)  that  will  be  circumscribed  by  the 
hemispheres  BDA ;  and  the  rectangles  DN,  Fq,  Gr,  Hs,  will  also  describe 
cylinders  that  will  circumscribe  the  cone  ICE.  Now,  it  may  be  demon- 
strated, as  was  done  of  the  prisms  inscribed  in  a  pyramid  (13.  3.  Sup.), 
that  the  sum  of  all  the  cylinders  described  within  the  hemisphere,  is  ex- 
ceeded by  the  hemisphere  by  a  solid  less  than  the  cylinder  generated  by 
the  rectangle  HB,  that  is,  by  a  solid  less  than  W,  for  the  cylinder  generated 
by  HB  is  less  than  W.  In  the  same  manner,  it  may  be  demonstrated, 
that  the  sum  of  the  cylinders  circumscribing  the  cone  ICE  is  greater  than 


OE  GEOMETRY.    BOOK  III. 


215 


the  cone  by  a  solid  less  than  the  cylinder  generated  by  the  rectangle  DN, 
that  is,  by  a  solid  less  than  W.  Therefore,  since  the  sum  of  the  cylinders 
inscribed  in  the  hemisphere,  together  with  a  solid  less  than  W,  is  equal  to 
the  hemisphere  ;  and,  since  the  sum  of  the  cylinders  described  about  the 
cone  is  equal  to  the  cone  together  with  a  solid  less  than  W  ;  adding  equals 
to  equals,  the  sum  of  all  these  cylinders,  together  with  a  solid  less  than  W, 
is  equal  to  the  sum  of  the  hemisphere  and  the  cone  together  with  a  solid 
less  than  W.  Therefore,  the  difference  between  the  whole  of  the  cylin- 
ders and  the  sum  of  the  hemisphere  and  the  cone,  is  equal  to  the  difference 
of  two  solids,  which  are  each  of  them  less  than  W  ;  but  this  difference 
must  also  be  less  than  W,  therefore  the  difference  between  the  two  series 
of  cylinders  and  the  sum  of  the  hemisphere  and  cone  is  less  than  the  given 
solid  W. 


PROP.  XX.    THEOR. 

The  same  things  being  supposed  as  in  the  last  proposition,  the  sum  of  all  the 
cylinders  inscribed  in  the  hemisphere,  and  described  about  the  cone,  is  equal 
to  a  cylinder,  having  the  same  base  and  altitude  with  the  hemisphere. 

Let  the  figure  BCD  be  constructed  as  before,  and  supposed  to  revolve 
About  CD  ;  the  cylinders  inscribed  in  the  hemisphere,  that  is,  the  cylinders 
described  by  the  revolution  of  the  rectangles  Hh,  Gg,  Ff,  together  with 
those  described  about  the  cone,  that  is,  the  cylinders  described  by  the  revo- 
lution of  the  rectangles  Hs,  Gr,  Fq,  and  DN  are  equal  to  the  cylinder  de- 
scribed by  the  revolution  of  the  rectangle  BD. 

Let  L  be  the  point  in  which  GO  meets  the  circle  ABD,  then,  because 
CGL  is  a  right  angle  if  CL  be  joined,  the  circles  described  with  the  dis- 
tances CG  and  GL  are  equal  to  the  circle  described  with  the  distance  CL 
(2.  Cor.  6.  1  Sup.)  or  GO ;  now,  CG  is  equal  to  GR,  because  CD  is  equal 
to  DE,  and  therefore  also,  the  circles  described  with  the  distance  GR  and 
GL  are  together  equal  to  the  circle  described  with  the  distance  GO,  that 
is,  the  circles  described  by  the  revolution  of  GR  and  GL  about  the  point 
G,  are  together  equal  to  the  circle  described  by  the  revolution  of  GO  about 
the  same  point  G ;  therefore  also,  the  cylinders  that  stand  upon  the  two 
first  of  these  circles,  having  the  common  altitudes  GH,  are  equal  to  the 


216 


SUPPLEMENT  TO  THE  ELEMENTS,  &c. 


cylinder  which  stands  on  the  remaining  circle,  and  which  has  the  same 
altitude  GH.  The  cylinders  described  by  the  revolution  of  the  rectangles 
Gg,  and  Gr  are  therefore  equal  to  the  cylinder  described  by  the  rectangle 
GP.  And  as  the  same  may  be  shewn  of  all  the  rest,  therefore  the  cylin- 
ders described  by  the  rectangles  Hh,  Gg,  Ff,  and  by  the  rectangles  Hs,  Gr, 
Fq,  DN,  are  together  equal  to  the  cylinder  described  by  BD,  that  is,  to  the 
cylinder  having  the  same  base  and  altitude  with  the  hemisphere. 

PROP.  XXL    THEOR. 

Every  sphere  is  two-thirds  of  the  circumscribing  cylinder. 

Let  the  figure  be  constructed  as  in  the  two  last  propositions,  and  if  the 
hemisphere  described  by  BDC  be  not  equal  to  two-thirds  of  the  cylinder 
described  by  BD,  let  it  be  greater  by  the  solid  W.  Then,  as  the  cone  de- 
scribed by  CDE  is  one-third  of  the  cylinder  (18.  3.  Sup.)  described  by  BD, 
the  co^e  and  the  hemisphere  together  will  exceed  the  cylinder  by  W.     But 


that  cylinder  is  equal  to  the  sum  of  all  the  cylinders  described  by  the  rect- 
angles Hh,  Gg,  Ff,  Hs,  Gr,  Fq,  DN  (20.  3.  Sup.) ;  therefore  the  hemisphere 
and  the  cone  added  together  exceed  the  sum  of  all  these  cylinders  by  the 
given  solid  W,  which  is  absurd  ;  for  it  has  been  shewn  (19.  3.  Sup.),  that 
the  hemisphere  and  the  cone  together  differ  from  the  sum  of  the  cylinders 
by  a  solid  less  than  W.  The  hemisphere  is  therefore  equal  to  two-thirds 
of  the  cylinder  described  by  the  rectangle  BD  ;  and  therefore  the  whole 
sphere  is  equal  to  two-thirds  of  the  cylinder  described  by  twice  the  rectan- 
gle BD,  that  is,  to  two-thirds  of  the  circumscribing  cylinder. 


END  OF  THE  SUPPLEMENT  TO  THE  ELEMENTS. 


ELEMENTS 


OF 


PLANE  TRIGONOMETRY. 


Trigonometry  is  the  application  of  Arithmetic  to  Geometry :  or,  more 
precisely,  it  is  the  application  of  number  to  express  the  relations  of  the  sides 
and  angles  of  triangles  to  one  another.  It  therefore  necessarily  supposes 
the  elementary  operations  of  arithmetic  to  be  understood,  and  it  borrows 
from  that  science  several  of  the  signs  or  characters  which  peculiarly  be- 
long to  it. 

The  elements  of  Plane  Trigonometry,  as  laid  down  here,  are  divided  into 
three  sections  :  the  first  explains  the  principles  ;  the  second  delivers  the 
rules  of  calculation ;  the  third  contains  the  construction  of  trigonometrical 
tables,  together  with  the  investigation  of  some  theorems,  useful  for  extend- 
ing trigonometry  to  the  solution  of  the  more  difficult  problems. 


SECTION   I. 


LEMMA  I. 

An  angle  at  the  centre  of  a  circle  is  to  four  right  angles  as  the  arc  on  which 
it  stands  is  to  the  whole  circumference* 

Let  ABC  be  an  angle  at  the  centre  of  the  circle  ACF,  standing  on  the 
circumference  AC  :  the  angle  ABC  is  to  four  right  angles  as  the  arc  AC 
to  the  whole  circumference  ACF. 

Produce  AB  till  it  meet  the  circle 
in  E,  and  draw  DBF  perpendicular  to 
AE. 

Then,  because  ABC,  ABD  are  two 
angles  at  the  centre  of  the  circle  ACF, 
the  angle  ABC  is  to  the  angle  ABD  as 
the  arc  AC  to  the  arc  AD,  (33.  6.) ; 
and  therefore  also,  the  angle  ABC  is  to 
four  times  the  angle  ABD  as  the  arc 
AC  to  four  times  the  arc  AD  (4.  5.). 

But  ABD  is  a  right  angle,  and  there- 
fore four  times  the  arc  AD  is  equal  to 

28 


218  PLANE  TRIGONOMETRY. 

the  whole  circumference  ACF  ;  therefore  the  angle  ABC  is  to  four  right 
angles  as  the  arc  AC  to  the  whole  circumference  ACF. 

CoR.  Equal  angles  at  the  centres  of  dilTerent  circles  stand  on  arcs  which 
have  the  same  ratio  to  their  circumferences.  For,  if  the  angle  ABC,  at 
the  centre  of  the  circles  ACE,  GHK,  stand  on  the  arcs  AC,  GH,  AC  is 
to  the  whole  circumference  of  the  circle  ACE,  as  the  angle  ABC  to  four 
right  angles  ;  and  the  arc  HG  is  to  the  whole  circumference  of  the  circle 
GHK  in  the  same  ratio. 


DEFINITIONS. 


1 .  If  two  straight  lines  intersect  one  another  in  the  centre  of  a  circle,  the 
arc  of  the  circumference  intercepted  between  them  is  called  the  Measure 
of  the  angle  which  they  contain.  Thus  the  arc  AC  is  the  measure  of 
the  angle  ABC. 

2.  If  the  circumference  of  a  circle  be  divided  into  360  equal  parts,  each  of 
these  parts  is  called  a  Degree ;  and  if  a  degree  be  divided  into  60  equal 
parts,  each  of  these  is  called  a  Minute  ;  and  if  a  minute  be  divided  into 
60  equal  parts,  each  of  them  is  called  a  Second^  and  so  on.  And  as  many 
degrees,  minutes,  seconds,  (fee.  as  are  in  any  arc,  so  many  degrees,  mi- 
nutes, seconds,  &c.  are  said  to  be  in  the  angle  measured  by  that  arc. 

Cor.  1.  Any  arc  is  to  the  whole  circumference  of  which  it  is  a  part,  as 
the  number  of  degrees,  and  parts  of  a  degree  contained  in  it  is  to  the 
number  360.  And  any  angle  is  to  four  right  angles  as  the  number  of 
degrees  and  parts  of  a  degree  in  the  arc,  which  is  the  measure  of  that 
angle,  is  to  360. 

CoR.  2.  Hence  also,  the  arcs  which  measure  the  same  angle,  whatever 
be  the  radii  with  which  they  are  described,  contain  the  same  number  of 
degrees,  and  parts  of  a  degree.  For  the  number  of  degrees  and  parts  of 
a  degree  contained  in  each  of  these  arcs  has  the  same  ratio  to  the  num- 
ber 360,  that  the  angle  which  they  measure  has  to  four  right  angles 
(Cor.  Lem.  1.). 

The  degrees,  minutes,  seconds,  &;c.  contained  in  any  arc  or  angle,  are 
usually  written  as  in  this  example,  49^.  36'.  24".  42'"  ;  that  is,  49  de- 
grees, 36  minutes,  24  seconds,  and  42  thirds. 

3.  Two  angles,  which  are  together  equal  to  two  right  angles,  or  two  arcs 
which  are  together  equal  to  a  semicircle,  arc  called  the  Supplements  of 
one  another. 

4.  A  straight  line  CD  drawn  through  C,  one  of  the  extremities  of  the  arc 


PLANE  TRIGONOMETRY. 


219 


AC,  perpendicular  to  the  diameter 
passing  through  the  other  extremity 
A,  is  called  the  Sine  of  the  arc  AC, 
or  of  the  angle  ABC,  of  which  AC 
is  the  measure. 

CoR.  1.  The  sine  of  a  quadrant,  or  of 
a  right  angle,  is  equal  to  the  radius. 

CoR.  2.  The  sine  of  an  arc  is  half  the 
chord  of  twice  that  arc  :  this  is  evi- 
dent by  producing  the  sine  of  any 
arc  till  it  cut  the  circumference. 


5.  The  segment  DA  of  the  diameter  passing  through  A,  one  extremity  of 
the  arc  AC,  between  the  sine  CD  and  the  point  A,  is  called  the  Versed 
sine  of  the  arc  AC,  or  of  the  angle  ABC, 

6.  A  straight  line  AE  touching  the  circle  at  A,  one  extremity  of  the  arc 
AC,  and  meeting  the  diameter  BC,  which  passes  through  C  the  other 
extremity,  is  called  the  Tangent  of  the  arc  AC,  or  of  the  angle  ABC 

Cor.     The  tangent  of  half  a  right  angle  is  equal  to  the  radius. 

7.  The  straight  line  BE,  between  the  centre  and  the  extremity  of  the  tan- 
gent AE  is  called  the  Secant  of  the  arc  AC,  or  of  the  angle  ABC. 

Cor.  to  Def.  4,  6, 7,  the  sine,  tangent  and  secant  of  any  angle  ABC,  are 
likewise  the  sine,  tangent,  and  secant  of  its  supplement  CBF. 

It  is  manifest,  from  Def.  4.  that  CD  is  the  sine  of  the  angle  CBF.  Let 
CB  be  produced  till  it  meet  the  circle  again  in  I ;  and  it  is  also  mani- 
fest, that  AE  is  the  tangent,  and  BE  the  secant,  of  the  angle  ABI,  or 
CBF,  from  Def.  6.  7. 

CoR.  to  Def.  4,  5,  6,  7.  The  sine,  versed  sine,  tangent,  and  secant  of  an 
arc,  which  is  the  measure  of  any  gi- 
ven angle  ABC,  is  to  the  sine,  versed 
sine,  tangent  and  secant,  of  any  other 
arc  which  is  the  measure  of  the  same 
angle,  as  the  radius  of  the  first  arc  is 
to  the  radius  of  the  second. 

J  jet  AC,  MN  be  measures  of  the  angle 

ABC,  according  to  Def.  1.;  CD  the       ^  mvrTfc 

sine,  DA  the  versed  sine.     AE  the  CJ  J51.Xy 

tangent,  and  BE  the  secant  of  the  arc  AC,  according  to  Def.  4,  5,  6,  7  , 
NO  the  sine,  OM  the  versed  sine,  MP  the  tangent,  and  BP  the  secant 
of  the  arc  MN.  according  to  the  same  definitions.  Since  CD,  NO,  AE, 
MP  are  parallel,  CD  :  NO  : :  rad.  CB  :  rad.  NB,  and  AE  :  MP  :  :  rad. 
AB  :  rad.  BM,  also  BE  :  BP  :  :  AB  :  BM  ;  likewise  because  BC  :  BD 
:  :  BN  :  BO,  that  is,  BA  :  BD  : :  BM  :  BO,  by  conversion  and  alterna- 
tion, AD  :  MO  : :  AB  :  MB.     Hence  the  corollary  is  manifest.     And 


220 


PLANE  TRIGONOMETRY. 


therefore,  if  tables  be  constructed,  exliibiting  in  numbers  the  sines,  tan- 
gents, secants,  and  versed  sines  of  certain  angles  to  a  given  radius,  they 
will  exhibit  the  ratios  of  the  sines,  tangents,  &c.  of  the  same  angles  to 
any  radius  whatsoever. 
In  such  tables,  which  are  called  Trigonometrical  Tables,  the  radius  is 
either  supposed  I,  or  some  in  the  series  10,  100,  1000;  &;c.  The  use 
and  construction  of  these  tables  are  about  to  be  explained. 

8.  The  difference  between  any  angle 
and  a  right  angle,  or  between  any 
arc  and  a  quadrant,  is  called  the 
Complement  of  that  angle,  or  of  that 
arc.  Thus,  if  BH  be  perpendicular 
to  AB,  the  angle  CBH  is  the  com- 
plement of  the  angle  ABC,  and  the 
arc  HO  the  complement  of  AC  ; 
also,  the  complement  of  the  obtuse 
angle  FBC  is  the  angle  HBC,  its 
excess  above  a  right  angle ;  and 
the  complement  of  the  arc  FC  is 
HC. 

9.  The  sine,  tangent,  or  secant  of  the  complement  of  any  angle  is  called 
the  Cosine,  Cotangent,  or  Cosecant  of  that  angle.  Thus,  let  CL  or  DB, 
which  is  equal  to  CL,  be  the  sine  of  the  angle  CBH  ;  HK  the  tangent, 
and  BK  the  secant  of  the  same  angle :  CL  or  BD  is  the  cosine,  HK  the 
cotangent,  and  BK  the  cosecant  of  the  angle  ABC. 

CoR.  1.  The  radius  is  a  mean  proportional  between  the  tangent  and  the 
cotangent  of  any  angle  ABC  ;  that  is,  tan.  ABC  X  cot.  ABC=R2. 

For,  since  HK,  BA  are  parallel,  the  angles  HKB,  ABC  are  equal,  and 
KHB,  BAE  are  right  angles ;  therefore  the  triangles  BAE,  KHB  are 
similar,  and  therefore  AE  is  to  AB,  as  BH  or  BA  to  HK. 

CoR.  2.     The  radius  is  a  mean  proportional  between  the  cosine  and  se- 
cant of  any  angle  ABC  ;  or 
COS.  ABC  X  sec.  ABC=R2. 

Since  CD,  AE  are  parallel,  BD  is  to  BC  or  BA,  as  BA  to  BE. 


PROP.  L 

In  a  right  angled  plane  triangle,  as  the  hypotenuse  to  either  of  the  sides,  so 
the  radius  to  the  sine  of  the  angle  opposite  to  that  side  ;  and  as  either  of  the 
sides  is  to  the  other  side,  so  is  the  radius  to  the  tangent  of  the  angle  oppo' 
site  to  that  side. 

Let  ABC  be  aright  angled  plane  triangle,  of  which  BC  is  the  hypote- 
nuse. From  the  centre  C,  with  any  radius  CD,  describe  the  arc  DE ; 
draw  DF  at  right  angles  to  CE,  and  from  E  draw  EG  touching  the  circle 
in  E,  and  meeting  CB  in  G  ;  DF  is  the  sine,  and  EG  the  tangent  of  the 
arc  DE,  or  of  the  angle  0. 


PLANE  TRIGONOMETRY. 


221 


10 


The  two  triangles  DFC,  BAG,  are  equiangular,  because  the  angles 
DFG,  BAG  are  right  angles,  and  the 
angle  at  G  is  common.  Therefore, 
CB  :  BA  :  :  GD  :  DF  ;  but  GD  is 
the  radius,  and  DF  the  sine  of  the 
angle  G,  (Def.  4.)  ;  therefore  GB  : 
BA  :  :  R  :  sin.  G. 

Also,  because  EG  touches  the  cir- 
cle in  E,  GEG  is  a  right  angle,  and 
therefore  equal  to  the  angle  BAG ; 
and  since  the  angle  at  G  is  common 
to  the  triangles  CBA,  GGE,  these  triangles  are  equiangular,  wherefore 
CA  :  AB  : :  GE  :  EG ;  but  GE  is  the  radius,  and  EG  the  tangent  of  the 
angle  G ;  therefore,  GA  :  AB  : :  R  :  tan.  G. 

GoR.  1.  As  the  radius  to  the  secant  of  the  angle  G,  so  is  the  side  adja- 
cent to  that  angle  to  the  hypotenuse.  For  GG  is  the  secant  of  the  angle 
G  (def.  7.),  and  the  triangles  GGE,  GBA  being  equiangular,  GA  :  GB  : : 
GE  :  GG,  that  is,  GA  :  GB  : :  R  :  sec.  G. 

Cor.  2.  If  the  analogies  in  this  proposition,  and  in  the  above  corollary 

be  arithmetically  expressed,  making  the  radius  =  1,  they  give  sin.  G  = 

AB     .       ^      AB     '^^       BG      ^ 
sec.  C  = 


BC' 


tan.  C 


AG' 


/f> 


AG* 


ffio", 


since  sin.  G=cos.  B,  because  B 


AB 


is  the  complement  of  G,  cos.  B  =k7^>  ^^<i  ^o^  the  same  reason,  cos.  C  = 

BO 

AC 
BC' 

Cor.  3.  In  every  triangle,  if  a  perpendicular  be  drawn  from  any  of  the 
angles  on  the  opposite  side,  the  segments  of 
that  side  are  to  one  another  as  the  tangents  of 
the  parts  into  which  the  opposite  angle  is  di- 
vided by  the  perpendicular.  For,  if  in  the  tri- 
angle ABC,  AD  be  drawn  perpendicular  to 
the  base  BG,  each  of  the  triangles  CAD,  ABD 
being  right  angled,  AD  ;  DC  : :  R  r'tan.  CAD, 
and  AD  :  DB  : :  R  :  tan.  DAB  ;  therefore,  ex 
aequo,  DC  :  DB  : :  tan.  CAD  :  tan.  BAD. 


SCHOLIUM. 

The  proposition,  just  demonstrated,  is  most  easily  remembered,  by  stating 
it  thus  :  If  in  a  right  angled  triangle  the  hypotenuse  be  made  the  radius, 
the  sides  become  the  sines  of  the  opposite  angles  ;  and  if  one  of  the  sides  be 
made  the  radius,  the  other  side  becomes  the  tangent  of  the  opposite  angle, 
and  the  hypotenuse  the  secant  of  it. 


222 


PLANE  TRIGONOMETRY 


PROP.  II.     THEOR. 

The  sides  of  a  plane  triangle  are  to  one  another  as  the  sines  of  the  opposite 

angles. 

From  A  any  angle  in  the  triangle  ABC, 
let  AD  be  drawn  perpendicular  to  BC. 
And  because  the  triangle  ABD  is  right 
angled  at  D,  AB  :  AD  : :  R  :  sin.  B  ;  and 
for  the  same  reason,  AC  :  AD  :  :  R  : 
sin.  C,  and  inversely,  AD  :  AC  :  :  sin. 
C  :  R ;  therefore,  ex  aequo  inversely,  AB 
:  AC  :  :  sin.  C  :  sin.  B.  In  the  same 
manner  it  may  be  demonstrated,  that  AB 
:  BC  :  :  sin.  C  :  sin.  A. 


PROP.  III.     THEOR. 

The  sum  of  the  sines  of  any  two  arcs  of  a  circle,  is  to  the  difference  of  their 
sines,  as  the  tangent  of  half  the  sum  of  the  arcs  to  the  tangent  of  half  their 
difference. 

Let  AB,  AC  be  two  arcs  of  a  circle  ABCD ;  let  E  be  the  centre,  and 
AEG  the  diameter  which  passes  through  A  ;  sin.  AC+sin.  AB  :  sin.  AC 
-sin.  AB  :  :  tan.  ^  (AC+AB)  :  tan.  I  (AC-AB). 

Draw  BF  parallel  to  AG,  meeting  the  circle  again  in  F.  Draw  BH 
and  CL  perpendicular  to  AE,  and  they  will  be  the  sines  of  the  arcs  AB 
and  AC  ;  produce  CL  till  it  meet  the  circle  again  in  D  ;  join  DF,  FC,  DE, 
EB,  EC,  DB. 

Now,  since  EL  from  the  centre  is  perpendicular  to  CD,  it  bisects  the 
line  CD  in  L  and  the  arc  CAD  in  A : 
DL  is  therefore  equal  to  LC,  or  to  the 
sine  of  the  arc  AC  ;  and  BH  or  LK 
being  the  -sine  of  AB,  DK  is  the  sum 
of  the  sines  of  the  arcs  AC  and  AB, 
and  CK  is  the  difference  of  their  sines ; 
DAB  also  is  the  sum  of  the  arcs  AC 
and  AB,  because  AD  is  equal  to  AC, 
and  BC  is  their  difference.  Now,  in 
the  triangle  DFC,  because  FK  is  per- 
pendicular to  DC,  (3.  cor.  1.),  DK  : 
KC  :  :  tan.  DFK  :  tan.  CFK ;  but 
tan.  DFK=tan.  J  arc.  BD,  because 
the  angle  DFK  (20.  3.)  is  the  half  of  DEB,  and  therefore  measured  by 
half  the  arc  DB.  For  the  same  reason,  tan.  CFK=tan.  J  arc.  BC  ;  and 
consequently,  DK  :  KC  : :  tan.  i  arc.  BD  :  tan.  i  arc.  BC.  But  DK  is 
the  sum  of  the  sines  of  the  arcs  AB  and  AC  ;  and  KC  is  the  difference  of 
the  sines ;  also  BD  is  the  sum  of  the  arcs  AB  and  AC,  and  BC  the  difle- 
rence  of  those  arcs 


PLANE  TRIGONOMETRY.  223 

Cor.  1.  Because  EL  is  the  cosine  of  AC,  and  EH  of  AB,  FK  is  the 
sum  of  these  cosines,  and  KB  their  difference  ;  for  FK=^FB+EL=EH 
+EL,  and  KB=LH  =  EH-EL.  Now,  FK  :  KB  :  :  tan.  FDK  :  tan. 
BDK ;  and  tan.  DFK=cotan.  FDK,  because  DFK  is  the  complement 
of  FDK ;  therefore,  FK  :  KB  :  :  cotan.  DFK  :  tan.  BDK,  that  is,  FK  : 
KB  :  :  cotan.  -J  arc.  DB  :  tan.  -J  arc.  BC.  The  sum  of  the  cosines  of  two 
arcs  is  therefore  to  the  difference  of  the  same  cosines  as  the  cotangent  of 
half  the  sum  of  the  arcs  to  the  tangent  of  half  their  difference. 

Cor.  2.  In  the  right  angled  triangle  FKD,  FK  :  KD  :  :  R  :  tan.  DFK; 
Now  FK=cos.  AB+cos.  AC,  KD=  sin.  AB+sin.  AC,  and  tan.  DFK= 
tan.  J  (AB+AC),  therefore  cos.  AB  +  cos.  AC  :  sin.  AB+sin.  AC  :  :  R  : 
tan.i(AB+AC). 

In  the  same  manner,  by  help  of  the  triangle  FKC,  it  may  be  shewn  that 
COS.  AB+cos.  AC  :  sin.  AC— sin.  AB  :  :  R  :  tan.  i(AC— AB). 

Cor.  3.  If  the  two  arcs  AB  and  AC  be  together  equal  to  90°,  the  tan- 
gent of  half  their  sum,  that  is,  of  45°,  is  equal  to  the  radius.  And  the  arc 
BC  being  the  excess  of  DC  above  DB,  or  above  90°,  the  half  of  the  arc 
BC  will  be  equal  to  the  excess  of  the  half  of  DC  above  the  half  of  DB,  that 
is,  to  the  excess  of  AC  above  45°  ;  therefore,  when  the  sum  of  two  arcs  is 
90°,  the  sum  of  the  sines  of  those  arcs  is  to  their  difference  as  the  radius  to 
the  tangent  of  the  dijQference  between  either  of  them  and  45°. 

PROP.  IV.    THEOR. 

The  sum  of  any  two  sides  of  a  triangle  is  to  their  difference,  as  the  tangent  of 
half  the  sum  of  the  angles  opposite  to  those  sides,  to  the  tangent  of  half  their 
difference. 

Let  ABC  be  any  plane  triangle  ; 
CA+AB  :  CA— AB  :  :  tan.  i  (B  +  C) :  tan.  J  (B— C). 
For  (2.)  CA  :  AB  :  :  sin.  B  :  sin.  C ; 
and  therefore  (E.  5.) 

GA+AB  :  CA— AB  :  :  sin.  B-f-sin.  C  :  sin.  B— sin.  C. 
But,  by  the  last,  sin.  B+sin.  C  :  sin.  B — sin.  C  :  : 
tan.  J  (B  +  C)  :  tan.  1  (B— C) ;  therefore  also,  (1 1.  5.) 
CA+AB  :  CA-AB" :  :  tan.  J  (B  +  C)  :  tan.  J  (B-C). 


224  PLANE  TRIGONOMETRY. 

Otherwise,  without  the  3d. 

Let  ABC  be  a  triangle  ;  the  sum  of  AB  and  AC  any  two  sides,  is  to  the 
difference  of  AB  and  AC  as  the  tangent  of  half  the  sum  of  the  angles  ACB 
and  ABC,  to  the  tangent  of  half  their  difference. 

About  the  centre  A  with  the  radius  AB,  the  greater  of  the  two  sides,  de- 
scribe a  circle  meeting  BC  produced  in  D,  and  AC  produced  in  E  and  F. 
Join  DA,  EB,  FB  ;  and  draw  FG  parallel  to  CB,  meeting  EB  in  G. 


Because  the  exterior  angle  E  AB  is  equal  to  the  two  interior  ABC,  ACB 
(32.  1.) :  and  the  angle  EFB,  at  the  circumference  is  equal  to  half  the  an- 
gle EAB  at  the  centre  (20.  3.) ;  therefore  EFB  is  half  the  sum  of  the  an- 
gles opposite  to  the  sides  AB  and  AC. 

Again,  the  exterior  angle  ACB  is  equal  to  the  two  interior  CAD,  ADC, 
and  therefore  CAD  is  the  difference  of  the  angles  ACB,  ADC,  that  is,  of 
ACB,  ABC,  for  ABC  is  equal  to  ADC.  Wherefore  also  DBF,  which  is 
the  half  of  CAD,  or  BFG,  which  is  equal  to  DBF,  is  half  the  difference  of 
the  angles  opposite  to  the  sides  AB,  AC. 

Now  because  the  angle  FBE,  in  a  semicircle  is  a  right  angle,  BE  is  the 
tangent  of  the  angle  EFB,  and  BG  the  tangent  of  the  angle  BFG  to  the 
radius  FB  ;  and  BE  is  therefore  to  BG  as  the  tangent  of  half  the  sum  of 
the  angles  ACB,  ABC  to  the  tangent  of  half  their  difference.  Also  CE  is 
the  sum  of  the  sides  of  the  triangle  ABC,  and  CF  their  difference  ;  and  be- 
cause BC  is  parallel  to  FG,  CE  :  CF  :  :  BE  ;  BG,  (2.  6.)  that  is,  the  sum 
of  the  two  sides  of  the  triangle  ABC  is  to  their  difference  as  the  tangent  of 
half  the  sum  of  the  angles  opposite  to  those  sides  to  the  tangent  of  half 
their  difference. 


PLANE  TRIGONOMETRY. 


225 


PROP.  V.    THEOR. 

If  a  perpendicular  he  drawn  from  any  angle  of  a  triangle  to  the  opposite  side^ 
or  base ;  the  sum  of  the  segments  of  the  base  is  to  the  sufn  of  the  other  two 
sides  of  the  triangle  as  the  difference  of  those  sides  to  the  difference  of  the 
segments  of  the  base. 

For  (K.  6.),  the  rectangle  under  the  sum  and  difference  of  the  segments 
of  the  base  is  equal  to  the  rectangle  under  the  sum  and  difference  of  the 
sides,  and  therefore  (16.  6.)  the  sum  of  the  segments  of  the  base  is  to  th^ 
sum  of  the  sides  as  the  difference  of  the  sides  to  the  difference  of  the  seg- 
ments of  the  base. 


PROP.  VI.    THEOR. 


In  any  triangle^  twice  the  rectangle  contained  by  any  two  sides  is  to  the  dif" 
ference  between  the  sum  of  the  squares  of  those  sides,  and  the  square  of  the 
base,  as  the  radius  to  the  cosine  of  the  angle  included  by  the  two  sides. 

Let  ABC  be  any  triangle,  2AB.BC  is 
to  the  difference  between  AB^-j-BC^  and 
AC^  as  radius  to  cos.  B. 

From  A  draw  AD  perpendicular  to  BC, 
and  (12.  and  13.  2.)  the  difference  be- 
tween the  sum  of  the  squares  of  AB  and 
BC,  and  the  square  on  AC  is  equal  to 
2BC.BD. 

But   BC.BA  :  BC.BD  :  :  BA  :  BD  :  :       _  ^ 

R  :  cos.  B,  therefore  also  2BC.BA  :  2BC.      B  D  C 

BD  :  :  R  :  cos.  B.     Now  2BC.BD  is  the  difference  between  AB^+BC^ 
and  AC^  therefore  twice  the  rectangle 

AB.BC  is  to  the  difference  between  jk 

AB2-I-BC2,  and   AC^  as  radius  to  the 
cosine  of  B. 

CoR.  If  the  radius  =1,  BD=BA 
Xcos.  B,  (1.),  and  2BC.BAxcos.  B 
=2BC.BD,  and  therefore  when  B  is 
acute,  2BC.BAXC0S.  B  =  BC^-f  BA2 
— AC2,  and  adding  AC2  to  both;  AC^ 
4-  2  cos.  B  X  BC.BA  =  BC2+  BA2  ; 
and  taking  2  cos.  Bx  BC.BA  from  both,  AC2=BC2— 2  cos.  Bx  BC.BA 
-f-BA2.     Wherefore  AC=  V(BC2— 2  cos.  B  xBC.BA-f-BA^). 

If  B  is  an  obtuse  angle,  it  is  shewn  in  the  same  way  that  AC= 
V(BC2-i-2  cos.  BxBC.BA-f-BA2). 

29 


226 


PLANE  TRIGONOMETRY. 


PROP.  VII.    THEOR. 

Four  times  the  rectangle  contained  hy  any  two  sides  of  a  triangle^  is  to  the 
rectangle  contained  by  two  straight  lines,  of  which  one  is  the  base  or  third 
side  of  the  triangle  increased  by  the  difference  of  the  two  sides,  and  the 
other  the  base  diminished  by  the  difference  of  the  same  sides,  as  the  square 
of  the  radius  to  the  square  of  the  sine  of  half  the  angle  included  between  the 
two  sides  of  the  triangle. 

Let  ABC  be  a  triangle  of  which  BC  is  the  base,  and  AB  the  greater  of 
the  two  sides  ;  4AB.AC  :  (BC+(AB-AC))  X  (BC-(AB-.AC)) : :  R2 
:  (sin.  i  BAC)2. 

Produce  the  side  AC  to  D,  so  that  AD=AB  ;  join  BD,  and  draw  AE, 


CF  at  right  angles  to  it ;  from  the  centre  C  with  the  radius  CD  describe 
the  seniicircle  GDH,  cutting  BD  in  K,  BC  in  G,  and  meeting  BC  pro- 
duced in  H. 

It  is  plain  that  CD  is  the  difference  of  the  sides,  and  therefore  that  BH  is 
the  base  increased,  and  BG  the  base  diminished  by  the  difference  of  the 
sides  ;  it  is  also  evident,  because  the  triangle  BAD  is  isosceles,  that  DE  is 
the  half  of  BD,  and  DF  is  the  half  of  DK,  wherefore  DE— DF=the  half 
of  BD— DK  (6.  5.),  that  is,  EF=^  BK.  And  because  AE  is  drawn  pa- 
rallel to  CF,  a  side  of  the  triangle  CFD,  AC :  AD  :  :  EF  :  ED,  (2.  6.) ; 
and  rectangles  of  the  same  altitude  being  as  their  bases  ACAD  :  AD^  : : 
EF.ED  :  ED2  (1.  6.),  and  therefore  4AC.AD  :  AD^  : :  4EF.ED  :  ED^,  or 
alternately,  4AC.AD  :  4EF.ED  : :  AD^  :  ED^. 

But  since  4EF=2BK'  4EF.ED=2BK.ED=2ED.BK=DB.BK== 
HB.BG  ;  therefore  4AC.AD  :  DB.BK  : :  AD2  :  ED^.  Now  AD  :  ED  : : 
R  :  sin.  EAC=sin.  i  BAC  (1.  Trig.)  and  AD^  :  ED^  : :  R2 :  (sin.  -|  BAC)2 : 
therefore,  (11.  5.)  4AC.AD  :  HB.BG  : :  R^  :  (sin.  J  BAC)2,  or  since  AB 
=AD,  4AC.AB  :  HB.BG  :  :  R2  :  (sin.  J  BAC)2.  Now  4AC.AB  is  four 
times  the  rectangle  contained  by  the  sides  of  the  triangle ;  HB.BG  is  that 
contained  by  BC  +  (AB-AC)  and  BC-(AB-AC). 


Cor.  Hence  2  -/ACAD  :  VHB.BG  : :  R  :  sin  J  BAC. 


PLANE  TRIGONOMETRY. 


227 


PROP.  VIII.    THEOR. 

Four  times  the  rectangle  contained  hy  any  two  sides  of  a  triangle,  is  to  the 
rectangle  contained  by  two  straight  lines,  of  which  one  is  the  sum  of  those 
sides  increased  hy  the  base  of  the  triangle,  and  the  other  the  sum  of  the  same 
sides  diminished  by  the  base,  as  the  square  of  the  radius  to  the  square  of 
the  cosine  of  half  the  angle  included  between  the  two  sides  of  the  triangle. 

Let  ABC  be  a  triangle,  of  which  BC  is  the  base,  and  AB  the  greater  of 
the  other  two  sides,  4AB.AC  :  (AB+AC-f  BC)  (AB+AC— BC)  :  :  R^: 
(cos.  JRAC)2. 

From  the  centre  C,  with  the  radius  CB,  describe  the  circle  BLM,  meet- 
ing AC,  produced,  in  L  and  M.  Produce  AL  to  N,  so  that  AN=AB  ;  let 
AD=AB  ;  draw  AE  perpendicular  to  BD ;  join  BN,  and  let  it  meet  the 
circle  again  in  P ;  let  CO  be  perpendicular  to  BN ;  and  let  it  meet  AE  in  R. 

It  is  evidentthat  MN=AB-|-AC+BC;  and  that  LN=AB4-AC— 

-BC.     Now,  because  BD  is  bisected  in  E,  and  DN  in  A,  BN  is  parallel  to 

AE,  and  is  therefore  perpendicular  to  BD,  and  the  triangles  DAE,  DNB 

are  equiangular;  wherefore,  since  DN=2AD.BN=2AE,  and  BP=2B0 

=2RE  ;  also  PN=2AR. 

But  because  the  triangles  ARC  and  AED  are  equiangular,  AC  :  AD  :  : 
AR  :  AE,  and  because  rectangles  of  the  same  altitude  are  as  their  bases 


B 


0/ 

A 

A^ 

\     ] 

s       ^ 

(1.  6.),  ACAD  :  AD2  : :  AR.AE  :  AE2,and  alternately  ACAD  :  AR  AE 
: :  AD2  :  AE2,  and  4ACAD  :  4AR.AE  : :  AD2  :  AE^.  But  4AR.AE= 
2ARx2AE=:NP.NB=:MN.NL  ;  therefore  4  AC  AD  :  MN  NL  •  •  AD^  • 

ffp*  An''\i?i(Tr'^^  -^^  '  '°'-  ^^^  (^)  ="°«-  2  (BACJ:  Wherefore 
4ACAD  :  MN.NL  : :  R2  :  (cos.  J  BAC)2 


^8  PLANE  TRIGONOMETRY. 

Now  4AC.AD  is  four  times  tlie  rectangle  under  the  sides  AC  and  AB, 
(for  AD=AB),  and  MN.NL  is  the  rectangle  under  the  sum  of  the  sides 
increased  by  the  base,  and  the  sum  of  the  sides  diminished  by  the  base. 

Cor.   1.     Hence  2  v'AC.AB  :  VMN.NL  : :  R  :  cos.  J  BAC. 

Cor.  2.  Since  by  Prop.  7.  4AC.AB  :  (BC+(AB-AC))  (BC-"(AB 
— BC))  : :  R2  :  (sin.  ^  BAC)^ ;  and  as  has  been  now  proved  4AC.AB  : 
(AB+AC+BC)  (AB  +  AC-BC)  :  :  R2  :  (cos.  }  BAC)2;  therefore,  ex 
aequo,  (AB  +  AC  +  BC)  (AB+AC-BC)  :  (BC  +  (AB-AC))  (BC- 
(AB— AC))  : :  (cos.  i  BAC)^  :  (sin.  J  BAC)^.  But  the  cosine  of  any  arc 
is  to  the  sine,  as  the  radius  to  the  tangent  of  the  same  arc  ;  therefore,  (AB 
+AC+BC)  (AB  +  AC-BC)  :  (BC+(AB-AC))  BC-(AB-.AC))  : : 
R2:  (tan.  ^BAC)2;  and 

■v/(AB4-AC  +  BC)  (AB+AC-BC  : 
V(BC+AB-AC)  (BC— (AB— AC))  : :  R  :  tan.  J  BAC. 


^' 


LEMMA  n. 

there  he  two  unequal  magnitudes^  half  their  difference  added  to  halfthetr 
sum  is  equal  to  the  greater ;  and  half  their  difference  taken  from  half  their 
sum  is  equal  to  the  less. 

Let  AB  and  BC  be  two  unequal  magnitudes,  of  which  AB  is  the  great- 
er ;  suppose  AC  bisected  in  D,  and  AE 

equal  to  BC.     It  is  manifest  that  AC  is     'X  E     D     B  C 

the  sum,  and  EB   the  difference  of  the 

magnitudes.  And  because  AC  is  bisected  in  D,  AD  is  equal  to  DC  :  but 
AE  is  also  equal  to  BC,  therefore  DE  is  equal  to  DB,  and  DE  or  DB  is 
half  the  difference  of  the  magnitudes.  But  AB  is  equal  to  BD  and  DA, 
that  is,  to  half  the  difference  added  to  half  the  sum  ;  and  BC  is  equal  to 
the  excess  of  DC,  half  the  sum  above  DB,  half  the  difference. 

Cor.  Hence,  if  the  sum  and  the  difference  of  two  magnitudes  be  given, 
the  magnitudes  themselves  may  be  found  ;  for  to  half  the  sum  add  half  the 
difference,  and  it  will  give  the  greater :  from  half  the  sum  subtract  half 
the  difference,  and  it  will  give  the  less. 

SCHOLIUM. 

This  property  is  evident  from  the  algebraical  sum  and  difference  of  the 
two  quantities  a  and  Z>,  of  which  a  is  the  greater ;  let  their  sum  be  denoted 
by  s,  and  their  difference  by  d :  then, 
a-\-b=s 


a-i-b^s  > 
a-b=dS 


.'.  by  addition,  2a=s-{-d; 

and  «=Y+-2- 
By  subtraction,  2b=:s — d ; 
and.. .5=---. 


r^^^ 

9KP 


PLANE  TRIGONOMETRY.  239 


SECTION  n. 


OF  THE  RULES  OF  TRIGONOMETRICAL 
CALCULATION. 

The  General  Problem  which  Trigonometry  proposes  to  resolve  is: 
In  any  'plane  triangle^  of  the  three  sides  and  the  three  angles,  any  three  being 
given,  and  one  of  these  three  being  a  side,  to  find  any  of  the  other  three. 

The  things  here  said  to  be  given  are  understood  to  be  expressed  by  their 
numerical  values  :  the  angles,  in  degrees,  minutes,  &c.;  and  the  sides  in 
feet,  or  any  other  known  measure. 

The  reason  of  the  restriction  in  this  problem  to  those  cases  in  which  at 
least  one  side  is  given,  is  evident  from  this,  that  by  the  angles  alone  being 
given,  the  magnitudes  of  the  sides  are  not  determined.  '  Innumerable  tri- 
angles, equiangular  to  one  another,  may  exist,  without  the  sides  of  any 
one  of  them  beiag  equal  to  those  of  any  other ;  though  the. ratios  of  their 
sides  to  one  another  will  be  the  same  in  them  all  (4.  6.).  If  therefore,  only 
the  three  angles  are  given,  nothing  can  be  determined  of  the  triangle  but 
the  ratios  of  the  sides,  which  may  be  found  by  trigonometry,  as  being  the 
same  with  the  ratios  of  the  sines  of  the  opposite  angles. 

For  the  conveniency  of  calculation,  it  is  usual  to  divide  the  general  pro- 
blem into  two  ;  according  as  the  triangle  has,  or  has  not,  one  of  the  angles 
a  right  angle. 

PROBLEM  I. 

In  a  right  angled  triangle,  of  the  three  sides,  and  three  angles,  any  two  being 
given,  besides  the  right  angle,  and  one^of  those  two  being  a  side,  it  is  required 
to  find  the  other  three. 

It  is  evident,  that  when  one  of  the  acute  angles  of  a  right  angled  triangle 
is  given,  the  other  is  given,  being  the  complement  of  the  former  to  a  right 
angle  ;  it  is  also  evident  that  the  sine  of  anyof  the  acute  angles  is  the 
cosine  of  the  other. 

This  problem  admits  of  several  cases,  and  the  solutions,  or  rules  for  cal- 
culation, which  all  depend  on  the  first  Proposition,  may  be  conveniently 
exhibited  in  the  form  of  a  table  ;  where  the  first  column  contains  the  things 
given  ;  the  second,  the  things  required ;  and  the  third,  the  rules  or  propo- 
sitions by  which  they  are  found. 


230 


PLANE  TRIGONOMETRY. 


GIVEN. 

SOOGHT. 

SOLUTION. 

CB  and  B,  the 
hypotenuse     and 
angle. 

AC. 
AB. 

R  :  sin  B  :  :  CB  :  AC. 
R  :  cos  B  : :  CB  :  AB. 

1 
2 

AC    and  C,  a 

side   and    one  of 
the  acute  angles. 

BC. 
AB. 

Cos  C  :  R  : :  AC  :  BC. 
R  :  tan  0  : :  AC  ;  AB. 

3 

4 

CB    and    BA, 

the      hypotenuse 
and  a  side. 

C 
AC. 

CB  :  BA  : :  R  :  sin  C. 
R  :  cos  C  :  :  CB  :  AC. 

5 
6 

AC    and    AB, 
the  two  sides. 

C. 
CB. 

AC  :  AB  : :  R  :  tan  C. 
Cos  C  :  R  : :  AC  :  CB. 

7 
8 

Remarks  on  the  Solutions  in  the  table. 


In  the  second  case,  when  AC  and  C  are  given  to  find  the  hypotenuse 
BC,  a  solution  may  also  be  obtained  by  help  of  the  secant,  for  CA  :  CB  :  : 
R  :  sec.  C. ;  if,  therefore,  this  proportion  be  made  R  :  sec.  C  :  :  AC  :  CB, 
CB  will  be  found.  , 

In  the  third  case,  when  the  hypotenuse  BC  and  the  side  AB  are  given 
to  find  AC,  this  may  be  done  either  as  directed  in  the  Table,  or  by  the 
47th  of  the  first;  for  since  AC2  =  BC2  ~  BA2,  AC  =  /BC^  -  BA2. 
This  value  of  AC  will  be  easy  to  calculate  by  logarithms,  if  the  quantity 
BC2 — BA2  be  separated  into  two  multipliers,  which Tnay  be  done  ;  because 
(Cor.  5.  2.),  BC2-BA2==(BC  +  BA)  .  (BC-BA).  Therefore  AC  = 
V(BC+BA)  (BC-BA). 

When  AC  and  AB  are  given,  BC  may  be  found  from  the  47th,  as  in  the 
preceding  instance,  for  BC=  ^BA^-I-AC^.  But  BA^+AC^  cannot  be 
separated  into  two  multipliers  ;  and  therefore,  when  BA  and  AC  are  large 
numbers,  this  rule  is  inconvenient  for  computation  by  logarithms.  It  is 
best  in  such  cases  to  seek  first  for  the  tangent  of  C,  by  the  analogy  in  the 
Table,  AC  :  AB  : :  R  :  tan.  C  ;  but  if  C  itself  is  notrequired,  itis  sufliicient, 
having  found  tan.  C  by  this  proportion,  to  take  from  the  Trigonometric 


PLANE  TRIGONOMETRY.  231 

Tables  the  cosine  that  corresponds  to  tan.  0,  and  then  to  compute  CB  from 
the  proportion  cos.  C  :  R  : :  AC  :  CB. 

PROBLExM  11. 

In  an  oblique  angled  triangle,  of  the  three  sides  and  three  angles,  any  three 
being  given,  and  one  of  these  three  being  a  side^  it  is  required  to  find  the 
other  three. 

This  problem  has  four  cases,  in  each  of  which  the  solution  depends  on 
some  of  the  foregoing  propositions. 

CASE  i. 

Tayo  angles  A  and  B,  and  one  side  AB,  of  a  triangle  ABC,  being  given, 
to  find  the  other  sides. 

SOLUTION. 

Because  the  angles  A  and  B  are  given,  C  is  also  given,  being  the  sup- 
plement of  A-j-B  ;  and,  (2.) 

Sin.  C  :  sin.  A  : :  AB  :  BC ;  also, 
Sin.  C  :  sin.  B  : :  AB  :  AC. 


CASE  II. 

Two  sides  AB  and  AC,  and  the  angle  B  opposite  to  one  of  them,  being 
given,  to  find  the  other  angles  A  and.C,  and  also  the  other  side  BC. 

SOLUTION. 

The  angle  C  is  found  from  this  proportion,  AC  :  AB  : :  sin.  B  :  sin.  C. 
Also,  A=180O— B— C  ;  and  then,  sin.  B  :  sin.  A  : :  AC  :  CB,  by  Case  1. 

In  this  case,  the  angle  C  may  have  two  values  ;  for  its  sine  being  found 
by  the  proportion  above,  the  angle  belonging  to  that  sine  may  either  be  that 
which  is  found  in  the  tables,  or  it  may  be  the  supplement  of  it  (Cor.  def.  4.). 
This  ambiguity,  however,  does  not  arise  from  any  defect  in  the  solution, 
but  from  a  circumstance  essential  to  the  problem,  viz.  that  whenever  AC 
is  less  than  AB,  there  are  two  triangles  which  have  the  sides  AB,  AC,  and 
the  angle  at  B  of  the  same  magnitude  in  each,  but  which  are  nevertheless 
unequal,  the  angle  opposite  to  AB  in  the  one,  being  the  supplement  of  that 
which  is  opposite  to  it  in  the  other.  The  truth  of  this  appears  by  describ- 
ing from  the  centre  A  with  the  radius  AC,  an  arc  intersecting  BC  in  C 


332  PLANE  TRIGONOMETRY. 

A. 


and  C  ;  then,  if  AC  and  AC  be  drawn,  it  is  evident  that  the  triangles 
ABC,  ABC  have  the  side  AB  and  the  angle  at  B  common,  and  the  sides 
AC  and  AC  equal,  but  have  not  the  remaining  side  of  the  one  equal  to  the 
remaining  side  of  the  other,  that  is,  BC  to  BC,  nor  their  other  angles  equal, 
viz.  BCA  to  BCA,  nor  BAC  to  13AC.  But  in  these  triangles  the  angles 
ACB,  ACB  are  the  supplements  of  one  another.  For  the  triangle  CAC 
is  isosceles,  and  the  angle  ACC=ACC,  and  therefore,  ACB,  which  is 
the  supplement  of  ACC,  is  also  the  supplement  of  ACC  or  ACB  ;  and 
these  two  angles,  ACB,  ACB  are  the  angles  found  by  the  computation 
above. 

From  these  two  angles,  the  two  angles  BAC,  BAC  will  be  found  :  the 
angle  BAC  is  the  supplement  of  the  two  angles  ACB,  ABC  (32.  1.),  and 
therefore  its  sine  is  the  same  with  the  sine  of  the  sum  of  ABC  and  ACB. 
But  BAC  is  the  difference  of  the  angles  ACB,  ABC  :  for  it  is  the  diffe- 
rence of  the  angles  ACC  and  ABC,  because  ACC,  that  is,  ACC  is  equal 
to  the  sum  of  the  angles  ABC,  BAC  (32.  1.).  Therefore,  to  find  BC, 
having  found  C,  make  sin.  C :  sin.  (C+B)  : :  AB  :  BC  ;  and  again,  sin. 
C  :  sin.  (C-B)  : :  AB  :  BC. 

Thus,  when  AB  is  greater  than  AC,  and  C  consequently  greater  than 
B,  there  are  two  triangles  which  satisfy  the  conditions  of  the  question. 
But  when  AC  is  greater  than  AB,  the  intersections  C  and  C  fall  on  oppo- 
site sides  of  B,  so  that  the  two  triangles  have  not  the  same  angle  at  B  com- 
mon to  them,  and  the  solution  ceases  to  be  ambiguous,  the  angle  required 
being  necessarily  less  than  B,  and  therefore  an  acute  angle. 

CASE  III. 

Two  sides  AB  and  AC,  and  the  angle  A,  between  them,  being  given  to 
find  the  other  angles  B  and  C,  and  also  the  side  BC. 

SOLUTION. 

First,  make  AB-fAC  :  AB— AC  : :  tan.  J  (C+B)  :  tan.  J  (C— B). 
Then,  since  ^  (C-j-B)  and  ^  (C— B)  are  both  given,  B  and  C  may  be  found. 
For  B=^  (C+B)+i  (C-B),  and  C=l  (C+B)-J  (C-B).    (Lem.  2.) 

To  find  BC. 


Having  found  B,  make  sin.  B  :  sin.  A  : :  AC  :  BC. 
But  BC  may  also  be  found  without  seeking  for  the  angle  B  and  C  j  for 
BC=  VAB2-2  COS.  AXAB.AC+AC2,  Prop.  6. 


PLANE  TRIGONOMETRY. 


233 


This  method  of  finding  BC  is  extremely  useful  in  many  geometrical  in- 
vestigations, but  it  is  not  very  well  adapted  for  computation  by  logarithms, 
because  the  quantity  under  the  radical  sign  cannot  be  separated  into  sim- 
ple multipliers.  Therefore,  when  AB  and  AC  are  expressed  by  large 
numbers,  the  other  solution,  by  finding  the  angles,  and  then  computing  BC, 
is  preferable. 

CASE  IV. 

The  three  sides  AB,  BC,  AC,  being  given,  to  find  the  angles  A,  B,  C. 

SOLUTION  I. 

Take  F  such  that  BC  :  BA+AG  : :  BA—AC  :  F,  then  F  is  either  the 
sum  or  the  difference  of  BD,  DC,  the  segments  of  the  base  (5.).  If  F  be 
greater  than  EC,  F  is  the  sum,  and  BC  the  difference  of  BD,  DC  ;  but,  if 
F  be  less  than  BC,  BC  is  the  sum,  and  F  the  difference  of  BD  and  DC. 
In  either  case,  the  sum  of  BD  and  DC,  and  their  difference  being  given, 
BD  and  DC  are  found.     (Lem.  2.) 

Then,  (1.)  CA  :  CD  : :  R  :  cos.  C  ;  and  BA  ;  BD  : :  R  ;  cos.  B  ;  where- 
fore C  and  B  are  given,  and  consequently  A. 


SOLUTION  II. 


Let  D  be  the  difference  of  the  sides  AB,  AC.  Then  (Cor.  7.)  2  VAB.AC 
V(BC+D)  (BC— D)  : :  R  :  sin.  J  BAG. 

SOLUTION  III. 


Let  S  be  the  sum  of  the  sides  BA  and  AC.   Then  (1,  Cor.  8.)  2  v'AB.AC 
V(S  +  BC)  (S~BC)  : :  R  :  cos.  i  BAG. 

SOLUTION  IV. 


S  and  D  retaining  the  significations  above,(2.Cor.8.)  •y/(S  +  BC)(S- 
VIBC  +  D)  (BG-D)  : :  R  :  tan.  J  BAG. 


BC) 


It  may  be  observed  of  these  four  solutions,  that  the  first  has  the  advan- 
tage of  being  easily  remembered,  but  that  the  others  are  rather  more  expe- 
ditious in  calculation.  The  second  solution  is  preferable  to  the  third,  when 
the  angle  sought  is  less  than  a  right  angle  ;  on  the  other  hand,  the  third 
is  preferable  to  the  second,  when  the  angle  sought  is  greater  than  a  right 

30 


234  PLANE  TRIGONOMETRY. 

angle  ;  and  in  extreme  cases,  that  is,  when  the  angle  sought  is  very  acute 
or  very  obtuse,  this  distinction  is  very  material  to  be  considered.  The 
reason  is,  that  the  sines  of  angles,  which  are  nearly  =  90°,  or  the  cosines 
of  angles,  which  are  nearly  =  0,  vary  very  little  for  a  considerable  varia- 
tion in  the  corresponding  angles,  as  maybe  seen  from  looking  into  the  ta- 
bles of  sines  and  cosines.  The  consequence  of  this  is,  that  when  the  sine 
or  cosine  of  such  an  angle  is  given  (that  is,  a  sine  or  cosine  nearly  equal  to 
the  radius,)  the  angle  itself  cannot  be  very  accurately  found.  If,  for  in- 
stance, the  natural  sine  .9998500  is  given,  it  will  be  immediately  per- 
ceived from  the  tables,  that  the  arc  corresponding  is  between  89°,  and  89° 
1' ;  but  it  cannot  be  found  true  to  seconds,  because  the  sines  of  89°  and  of 
89°  r,  differ  only  by  50  (in  the  two  last  places,)  whereas  the  arcs  them- 
selves differ  by  60  seconds.  Two  arcs,  therefore,  that  differ  by  1",  or  even 
by  more  than  1",  have  the  same  sine  in  the  tables,  if  they  fall  in  the  last 
degree  of  the  quadrant. 

The  fourth  solution,  which  finds  the  angle  from  its  tangent,  is  not  liable 
to  this  objection  ;  nevertheless,  when  an  arc  approaches  very  near  to  90°, 
the  variations  of  the  tangents  become  excessive,  and  are  too  irregular  to 
allow  the  proportional  parts  to  be  found  with  exactness,  so  that  when  the 
angle  sought  is  extremely  obtuse,  and  its  half  of  consequence  very  near  to 
90,  the  third  solution  is  the  best. 

It  may  always  be  known,  whether  the  angle  sought  is  greater  oif  less 
than  a  right  angle  by  the  square  of  the  side  opposite  to  it  being  greater  oi 
less  than  the  squares  of  the  other  two  sides. 


SECTION  III. 

CONSTRUCTION  OF  TRIGONOMETRICAL  TABLES. 

In  all  the  calculations  performed  by  the  preceding  rules,  tables  of  sines 
and  tangents  are  necessarily  employed,  the  construction  of  which  remains 
to  be  explained. 

The  tables  usually  contain  the  sines,  &c.  to  every  minute  of  the  quad- 
rant from  1'  to  90°,  and  the  first  thing  required  to  be  done,  is  to  compute 
the  sine  of  1',  or  of  the  least  arc  in  the  tables. 

1.  If  ADB  be  a  circle,  of  which  the  centre  is  C,  DB,  any  arc  of  that  cir- 
cle, and  the  arc  DBE  double  of  DB  ;  and  if  the  chords  DE,  DB  be  drawn, 
also  the  perpendiculars  to  them  from  C,  viz.  CF,  CG,  it  has  been  demon- 
strated (8.  1.  Sup.),  that  CG  is  a  mean  proportional  between  AH,  half  the 
radius,  and  AF,  the  line  made  up  of  the  radius  and  the  perpendicular  CF. 
Now  CF  is  the  cosine  of  the  arc  BD,  and  CG  the  cosine  of  the  half  of  BD  ; 
whence  the  cosine  of  the  half  of  any  arc  BD,  of  a  circle  of  which  the  ra- 
dius =  1,  is  a  mean  proportional  between  J  and  l-J-cos.  BD.  Or,  for  the 
greater  generality,  supposing  A  =  any  arc,  cos.  ^  A  is  a  mean  proportional 


PLANE  TRIGONOMETRY. 


235 


between  J  and  1 +cos.  A,  and  therefore  (cos.  ^  A)2=J  (i  -f-cos.  A)  or  cos. 
^A  =  Vi(l+cos.A). 

2.  From  this  theorem,  (which  is  the  same  that  is  demonstrated  (8. 1. 
Sup.),  only  that  it  is  here  expressed  trigonometrically,)  it  is  evident,  that  if 
the  cosine  of  any  arc  be  given,  the  cosine  of  half  that  arc  may  be  found. 
Let  BD,  therefore,  be  equal  to  60^,  so  that  the  chord  BD=radius,  then  the 
cosine  or  perpendicular  OF  was  shewn  (9.  L  Sup.)  to  be  =J,  and  there- 
fore cos.  J  BD,  or  COS.  30°=  v'J(l+i)==  V-4=~^-    ^^  ^^®  ^^^^  "^^^" 

ner,  cos.  150=  ^^(l+cos.SOo),  and  cos.  7°,  30'=  Vi(l  +  cos.l50),&c. 
In  this  way  the  cosine  of  3°,  45',  of  1°,  52',  30",  and  so  on,  will  be  com- 
puted, till  after  twelve  bisections  of  the  arc  of  60°,  the  cosine  of  52".  44'". 
93"".  45^.  is  found.  But  from  the  cosine  of  an  arc  its  sine  may  be 
found,  for  if  from  the  square  of  the  radius,  that  is,  from  1,  the  square  of 
the  cosine  be  taken  away,  ihe  remainder  is  the  square  of  the  sine,  and  its 
square  root  is  the  sine  itself.  Thus  the  sine  of  52".  44"'.  03"".  45^.  is 
found. 


3.  But  it  is  manifest,  that  the  sines  of  very  small  arcs  are  to  one  another 
nearly  as  the  arcs  themselves.  For  it  has  been  shewn  that  the  number  of 
the  sides  of  an  equilateral  polygon  inscribed  in  a  circle  may  be  so  great, 
that  the  perimeter  of  the  polygon  and  the  circumference  of  the  circle  may 
differ  by  a  line  less  than  any  given  line,  or,  which  is  the  same,  may  be 
nearly  to  one  another  in  the  ratio  of  equality.  Therefore  their  like  parts 
will  also  be  nearly  in  the  ratio  of  equality,  so  that  the  side  of  the  polygon 
will  be  to  the  arc  which  it  subtends  nearly  in  the  ratio  of  equality ;  and 
therefore,  half  the  side  of  the  polygon  to  half  the  arc  subtended  by  it,  that 
is  to  say,  the  sine  of  any  very  small  arc  will  be  to  the  arc  itself,  nearly  in 
the  ratio  of  equality..  Therefore,  if  two  arcs  are  both  very  small,  the  first 
will  be  to  the  second  as  the  sine  of  the  first  to  the  sine  of  the  second. 
Hence,  from  the  sine  of  52".  54'".  03"".  45^.  being  found,  the  sine  of  1' 


236 


PLANE  TRIGONOMETRY. 


becomes  known  ,  for,  as  52".  44'".  03"".  45^.  to  l,so  is  the  sine  of  the 
former  arc  to  the  sine  of  the  latter.  Thus  the  sine  of  1'  is  found  = 
0.0002908882. 

4.  The  sine  V  being  thus  found,  the  sines  of  2',  of  3',  or  of  any  number 
of  minutes,  may  be  found  by  the  following  proposition. 

THEOREM. 

Let  AB,  AC,  AD  be  three  such  arcs,  that  BC  the  difference  of  the  first 
and  second  is  equal  to  CD  the  difference  of  the  second  and  third  ;  the  ra- 
dius is  to  the  cosine  of  the  common  difference  BC  as  the  sine  of  AC,  the 
middle  arc,  to  half  the  sum  of  the  sines  of  AB  and  AD,  the  extreme  arcs. 

Draw  CE  to  the  centre  :  let  BF,  CG,  and  DH  perpendicular  to  AE,  be 
the  sines  of  the  arcs  AB,  AC,  AD.  Join  BD,  and  let  it  meet  CE  in  I ; 
draw  IK  perpendicular  to  AE,  also  BL  and 
IM  perpendicular  to  DH.  Then,  because 
the  arc  BD  is  bisected  in  C,  EC  is  at  right 
angles  to  BD,  and  bisects  it  in  I ;  also  BI  is 
the  sine,  and  EI  the  cosine  of  BC  or  CD. 
And,  since  BD  is  bisected  in  I,  and  IM  is 
parallel  to  BL  (2.  6.),  LD  is  also  bisected,  in 
M.  Now  BF  is  equal  to  HL,  therefore  BF 
4-DH=DH+HL  =  DL+2LH  =  2LM+ 
2LH=2MH  or  2KI ;  and  therefore  IK  is 
half  the  sum  of  BF  and  DH.  But  because 
the  triangles  CGE,  IKE  are  equiangular, 
CE  :  EI  :  :  CG  :  IK,  and  it  has  been  shewn  that  EI=cos.  BC,  and  IK= 
I  (BF+DH) ;  therefore  R  :  cos.  BC  :  :  sin.  AC  :  J  (sin.  AB+sin.  AD). 

Cor.     Hence,  if  the  point  B  coincide  with  A, 
R  :  COS.  BC  :  :  sin.  BC  :  J  sin.  BD,  that  is,  the  radius  is  to  the  cosine  of 
any  arc  as  the  sine  of  the  arc  is  to  half  the  sine  of  twice  the  arc  ;  or  if  any 
arc=A,  ^  sin.  2A=sin.  Axcos.  A,  or  sin.  2A=2  sin.  Ax  cos.  A. 

Therefore  also,  sin.  2' =2'  sin.  1'  x  cos.  1' :  so  that  from  the  sine  and 
cosine  of  one  minute  the  sine  of  2'  is  found. 

Again,  1',  2',  3',  being  three  such  arcs  that  the  difference  between  the 
first  and  second  is  the  same  as  between  the  second  and  third,  R  :  cos.  1'.: : 
sin.  2  :  ^  (sin.  I'+sin.  3'),  or  sin.  I'-j-sin.  3'=2  cos.  I'+sin.  2', and  taking 
sin.  1'  from  both,  sin.  3'=2  cos.  I'Xsin.  2'— sin.  1. 

In  like  manner,  sin.  4'=2'  cos.  I'X'sin.  3'— sin.  2, 
sin.  5'=2'  COS.  I'xsin.  4'— sin.  3, 
sin.  G'=2'  COS.  I'xsin.  5'— sin.  4,  &c. 

Thus  a  table  containing  the  sines  for  every  minute  of  the  quadrant  may 
be  computed ;  and  as  the  multiplier,  cos.  1'  remains  always  the  same,  the 
calculation  is  easy. 

For  computing  the  sines  of  arcs  that  differ  by  more  than  1',  the  method 
is  the  same.  Let  A,  A-fB,  A4-2B  be  three  such  arcs,  then,  by  this  the- 
orem, R  :  cos.B  : :  sin.  (A+B)  :  J  (sin.  A+sin.  (A-|-2B)) ;  .and  therefore 
making  the  radius,!. 


PLANE  TRIGONOMETRY.  237 

sin.  A+sin.  (A+2B)=2  cos.  Bxsin.  (A+B), 

orsin.  (A-i-2B)=2  COS.  Bxsin.  (A+B)— sin.  A. 
By  means  of  these  theorems,  a  table  of  the  sines,  and  consequently  also 
of  the  cosines,  of  arcs  of  any  number  of  degrees  and  minutes,  from  0  to  90, 

may  be  constructed.     Then,  because  tan.  A=  — '—^j  the  table  of  tangents 

cos.  A. 

is  computed  by  dividing  the  sine  of  any  arc  by  the  cosine  of  the  same  arc. 
When  the  tangents  have  been  found  in  this  manner  as  far  as  45°,  the  tan- 
gents for  the  other  half  of  the  quadrant  may  be  found  more  easily  by  an- 
other rule.  For  the  tangent  of  an  arc  above  45°  being  the  co-tangent  of 
an  arc  as  much  under  45°  ;  and  the  radius  being  a  mean  proportional  be- 
tween the  tangent  and  co-tangent  of  any  arc  (1.  Cor.  def.  9),  it  follows,  if 
the  difference  between  any  arc  and  45°  be  called  D,  that  tan.  (45° — D)  : 

1  : :  1  :  tan.  (450+D),  so  that  tan.  (450+J))=--Jg^-^^. 

Lastly,  the  secants  are  calculated  from  (Cor.  2.  def.  9.)  where  it  is 
shewn  that  the  radius  is  a  mean  proportional  between  the  cosine  and  the 

secant  of  any  arc,  so  that  if  Abe  anv  arc,  sec.  A= -. 

•^        '  >        '  cos.  A 

The  versed  sines  are  found  by  subtracting  the  cosines  from  the  radius. 

5.  The  preceding  Theorem  is  one  of  four,  which,  when  arithmetically 
expressed,  are  frequently  used  in  the  application  of  trigonometry  to  the  so- 
lution of  problems. 

Imo,  If  in  the  last  Theorem,  the  arc  AC=A,  the  arc  BC=B,  and  the 
radius  EC=I,  then  AD = A+B,  and  AB=A— B  ;  and  by  what  has  just 
been  demonstrated, 

1  :  cos.  B  : :  sin.  A  :  J  sin.  (A-f  B)-f-i  sin.  (A— B), 

and  therefore, 
sin.  Axcos.  B=lsin.  (A-f  B)+i  (A— B). 
2do,  Because  BF,  IK,  DH  are  parallel,  the  straight  lines  BD  and  FH 
are  cut  proportionally,  and  therefore  FH,  the  difference  of  the  straight  lines 
FE  and  HE,  is  bisected  in  K ;  and  therefore,  as  was  shewn  in  the  last 
Theorem,  KE  is  half  the  smn  of  FE  and  HE,  that  is,  of  the  cosines  of  the 
ares  AB  and  AD.     But  because  of  the  similar  triangles  EGC,  EKI,  EC 
;  EI  :  :  GE  :  EK  ;  now,  GE  is  the  cosine  of  AC,  therefore, 
R  :  cos.  BC  : :  cos.  AC  :  J  cos.  AD-fi  cos.  AB, 
or  I  :  cos.  B  : :  cos.  A  :  i  cos.  (A4-B)+ J  cos.  (A— B) ; 
and  therefore, 
cos.  Axcos.  B=i  COS.  (A-hB)+i  cos.  (A— B) ; 
3tio,  Again,  the  triangles  I  DM,  CEG  are  equiangular,  for  the  angles 
KIM,  EID  are  equal,  being  each  of  them  right  angles,  and  therefore,  tak- 
ing away  the  angle  EIM,  the  angle  DIM  is  equal  to  the  angle  EIK,  that 
is,  to  the  angle  ECG ;  and  the  angles  DMI,  CGE  are  also  equal,  being 
both  right  angles,  and  therefore  the  triangles  IDM,  CGE  have  the  sides 
about  their  equal  angles  proportionals,  and  consequently,  EC  :  CG  :  :  DI 
:  IM  ;  now,  IM  is  half  the  difference  of  the  cosines  FE  and  EH,  therefore, 
R  :  sin.  AC  :  :  sin.  BC  :  J  cos.  AB— i  cos.  AD, 
or  1  :  sin.  A  :  :  sin.  B  :  J  cos.  (A— B)— J  cos.  (A-f-B) ; 


238  PLANE  TRIGONOMETRY 

and  also, 
sin.  Ax  sin.  B=J  cos.  (A— B)— i  cos.  (A+B). 
4tOf  Lastly,  in  the  same  triangles  EGG,  DIM,  EC  :  EG  :  :  ID  :  DM ; 
now,  DM  is  half  the  difference  of  the  sines  DH  and  BE,  therefore, 
R  :  cos.  AC  :  :  sin.  BC  :  ^  sin.  AD— ^  sin.  AB, 
or  1  :  COS.  A  : :  sin.  B  :  J  sin.  (A+B)— J  sin.  (A+B) ; 
and  therefore, 
COS.  Ax  sin.  B=J  sin.  (A+B)— J  sin.  (A— B). 

6.  If  therefore  A  and  B  be  any  two  arcs  whatsoever,  the  radius  being 
supposed  1 ; 

I.  sin.  Ax  cos  B=Jsin.  (A+B)+isin.  (A— B). 
II.  cos.Axcos.  B=Jcos.  (A— B)-ficos.(A4-B) 

III.  sin.  Axsin.  B=|cos.(A-  B)— icos.(A-f  B). 

IV.  cos.  Axsin.  B=|sin.  (A+B)— } sin.  (A     B). 
From  these  four  Theorems  are  also  deduced  other  four. 

For  adding  the  first  and  fourth  together, 
sin.  Ax  COS.  B+cos.  Axsin.  B=sin.  (A+B). 

Also,  by  taking  the  fourth  /rom  the  first, 
sin.  Ax  COS.  B— cos.  Axsin.  B=sin.  (A— B). 
Again,  adding  the  second  and  third, 
COS.  Ax  cos.  B+sin.  Axsin.  B=cos.  (A— B) ; 
And,  lastly,  subtracting  the  third  from  the  second, 
cos.  Ax  COS.  B— sin.  Axsin.  B=cos.  (A+B). 

7.  Again,  since  by  the  first  of  the  above  theorems, 

sin.  Ax  COS. B=isin.(A+B)+J  sin.  (A-B),ifA+B=S,  and  A— B=D, 

v       /T         ON   A      S+D       ,^     S-D       ,       .        .     S+D 
then  (Lem.  2.)  A= — - — ,  and  B=-^- — ;  wherefore  sm.  — - —  X  cos. 

— - — =:Jsin.  S+JD.     But  as  S  and  D  maybe  any  arcs  whatever,  to 

preserve  the  former  notation,  they  may  be  called  A  and  B,  which  also  ex- 
press any  arcs  whatever :  thus, 

.     A+B  A— B     ,    .      ,  .  ,    .     „ 

sm.  — - — xcos.  — - — =i  sin.  A+ J  sm.  B,  or 

A.+B  A— B 

2  sin.  ^— - — Xcos. — - — =sin.  A+sin.  B. 

In  the  same  manner,  from  Theor.  2  is  derived, 

A+B  A— B  „  .  ,      ^         ,     ^, 

2  COS.  — - — X  COS.  — - — =cos.  B+cos.  A.     From  the  3d, 

^    .     A+B      .     A— B  ^  .  ,^         ^    ^^ 

2  sm.  — — -Xsm.  — - — =cos.  B— cos.  A ;  and  from  the  4tn, 

A+B       .     A-B      .      ,       .     „ 
2  COS. — - — Xsm.  — - — =sm.  A— sm.  B. 
2i  Z 

In  all  these  Theorems,  the  arc  B  is  supposed  less  than  A. 

8.  Theorems  of  the  same  kind  with  respect  to  the  tangents  of  arcs  may 
be  deduced  from  the  preceding.  Because  the  tangent  of  any  arc  is  equal 
to  the  sine  of  the  arc  divided  by  its  cosine. 


PLANE  TRIGONOMETRY.  239 

tan.  (A4-B)=  ^^"'  \        ..!•     But  it  has  just  been  shewn,  that 
^  ^      COS.  (A+B) 

sin.  (A+B)=sin.  Ax  cos.  B  +  cos.  Axsin.  B,  and  that 

COS.  (A+B)=cos.  Axcos.  B— sin.  Axsin.  B  ;  therefore,  tan.  (A+B)  = 

sin.  Ax  cos.  B+cos.  Axsin.  B       ,  j.  .,.     ,    .i    v  ^         jj 

—-^ — : : — — ,  and  dividing  both  the  numerator  and  deno- 

cos.  Ax  COS.  B— -sin.  Axsm.  B 

^  / ,      T^x      —tan.  A 4- tan.  B 

mmator  of  this  fraction  by  cos.  Axcos.  B,  tan.  (A+B)=— — r— — 5. 

i  tan.  A  X  tan.  £> 

TIM  /  *      Ti\         tan.  A  tan.  B 

In  hke  manner,  tan.  (A — B)=-—; 7,. 

'         ^  '     1+tan.  Axtan.  B 

9.  If  the  Theorem  demonstrated  in  Prop.  3,  be  expressed  in  the  same 
manner  with  those  above,  it  gives 

sin.  A+sin.  B    _  tan.  i  (A+B) 

sin.  A — sin.  B  ~  tan.  i  (A— B)* 
Also  by  Cor.  1,  to  the  3d, 

cos.  A+cos.  B  _    cot.  i  (A-4-B) 

cos.  A— cos.  B  ~~  tan.  ^  (A  — B)* 

And  by  Cor.  2,  to  the  same  proposition, 

sin,  A+sin.  B       tan.  i  (A+B)  .        ^   .    ,  1       , 

-— —  =  £.-^j or  since  R  is  here  supposed  =  1, 

cos.  A+cos.  B  R  '  ri  I 

sin.  A+sin.  B       ^       ,  .  .  ,  ^. 

cos.  A+cos.  B  2\     I     / 

10.  In  all  the  preceding  Theorems,  R,  the  radius,  isr  supposed  =1,  be- 
cause in  this  way  the  propositions  are  most  concisely  expressed,  and  are 
also  most  readily  applied  to  trigonometrical  circulation.  But  if  it  be  re- 
quired to  enunciate  any  of  them  geometrically,  the  multiplier  R,  which 
has  disappeared,  by  being  made  =  1,  must  be  restored,  and  it  will  always 
be  evident  from  inspection  in  what  terms  this  multiplier  is  wanting.  Thus, 
Theor.  1,2  sin.  A  X  cos.  B=sin.  (x\+B)+sin.  (A— B),  is  a  true  proposition, 
taken  arithmetically  ;  but  taken  geometrically,  is  absurd,  unless  we  sup- 
ply the  radius  as  a  multiplier  of  the  terms  on  the  right  hand  of  the  sine  of 
equality.  Itthenbecomes  2  sin.  Axcos.  B=R(sin.  (A+B)+sin.  (A— B)); 
or  twice  the  rectangle  under  the  sine  of  A,  and  the  cosine  of  B  equal  to  the 
rectangle  under  the  radius,  and  the  sum  of  the  sines  of  A+B  and  A — B. 

In  general,  the  number  of  li7iear  multipliers,  that  is,  of  lines  whose  nume- 
rical values  are  multiplied  together,  must  be  the  same  in  every  term,  other- 
wise we  will  compare  unlike  magnitudes  with  one  another. 

The  propositions  in  this  section  are  useful  in  many  of  the  higher  branches 
of  the  Mathematics,  and  are  the  foundation  of  what  is  called  the  Arithmetic 
of  Sines. 


ELEMENTS 


OF 


SPHERICAL 

TRIGONOMETRY. 


PROP.  I. 

If  a  sphere  he  cuthy  a  plane  through  the  centre^  the  section  is  a  circle,  having  the 
same  centre  with  the  sphere,  and  equal  to  the  circle  hy  the  revolutio7i  of  which 
the  sphere  was  described. 

For  all  the  straight  lines  drawn  from  the  centre  to  the  superficies  of  the 
sphere  are  equal  to  the  radius  of  the  generating  semicircle,  (Def.  7.  3. 
Sup.).  Therefore  the  common  section  of  the  spherical  superficies,  and  of 
a  plane  passing  through  its  centre,  is  a  line,  lying  in  one  plane,  and  hav- 
ing all  its  points  equally  distant  from  the  centre  of  the  sphere  ;  therefore  it 
is  the  circumference  of  a  circle  (Def.  11.  1.),  having  for  its  centre  the  cen- 
tre of  the  sphere,  and  for  its  radius  the.  radius  of  the  sphere,  that  is,  of  the 
semicircle  by  which  the  sphere  has  been  described.  It  is  equal,  therefore, 
to  the  circle  of  which  that  semicircle  was  a  part. 


DEFINITIONS. 


1.  Any  circle,  which  is  a  section  of  a  sphere  by  a  plane  through  its  centre, 
is  called  a  great  circle  of  the  sphere. 

CoR.  All  great  circles  of  a  sphere  are  ^qual ;  and  any  two  of  them  bisect 
one  another. 

They  are  all  equal,  having  all  the  same  radii,  as  has  just  been  shewn  ;  and 
any  two  of  them  bisect  one  another,  for  as  they  have  the  same  centre, 
their  common  section  is  a  diameter  of  both,  and  therefore  bisects  both. 

2.  The  pole  of  a  great  circle  of  a  sphere  is  a  point  in  the  superficies  of  the 
sphere,  from  which  all  strai  ^ht  lines  drawn  to  the  circumference  of  the 
circle  are  equal. 

3.  A  spherical  angle  is  an  angle  on  the  superficies  of  a  sphere,  contained 
by  the  arcs  of  two  great  circles  which  intersect  one  another ;  and  is  the 
same  with  the  inclination  of  the  planes  of  these  great  circles. 


SPHERICAL  TRIGONOMETRY. 


241 


4.  A  spherical  triangle  is  a  figure,  upon  the  superficies  of  a  sphere,  com- 
prehended by  three  arcs  of  three  great  circles,  each  of  which  is  less  than 


a  semicircle 


PROP.  II. 


The  arc  of  a  great  circle^  between  the  pole  and  the  circumference  of  another 
great  circle,  is  a  quadrant. 

Let  ABC  be  a  great  circle,  and  D  its  pole ;  if  DC,  an  arc  of  a  great 
circle,  pass  through  D,  and  meet  ABC  in  C,  the  ar6  DC  is  a  quadrant. 

Let  the  circle,  of  which  CD  is  an  arc,  meet  ABC  again  in  A,  and  let 
AC  be  the  common  section  of  the  planes 
of  these  great  circles,  which  will  pass 
through  E,  the  centre  of  the  sphere  :  Join 
DA,  DC.  Because  AD=DC,  (Def.  2.), 
and  equal  straight  lines,  in  the  same  cir- 
cle, cut  off  equal  arcs  (28.  3.),  the  arc  AD 
=  the  arc  DC  ;  but  ADC  is  a  semicircle, 
therefore  the  arcs  AD,  DC  are  each  of 
them  quadrants. 

CoR.  1.  If  DE  be  drawn,  the  angle  AED  is  a  right  angle  ;  and  DE 
being  therefore  at  right  angles  to  every  line  it  meets  with  in  the  plane  of 
the  circle  ABC,  is  at  right  angles  to  that  plane  (4.  2.  Sup.).  Therefore 
the  straight  line  drawn  from  the  pole  of  any  great  circle  to  the  centre  of  the 
sphere  is  at  right  angles  to  the  plane  of  that  circle ;  and,  conversely,  a 
straight  line  drawn  from  the  centre  of  the  sphere  perpendicular  to  the  plane 
of  any  greater  circle,  meets  the  superficies  of  the  sphere  in  the  pole  of  that 
circle. 

CoR.  2.  The  circle  ABC  has  two  poles,  one  on  each  side  of  its  plane, 
which  are  the  extremities  of  a  diameter  of  the  sphere  perpendicular  to  the 
plane  ABC  ;  and  no  other  points  but  these  two  can  be  poles  of  the  circle 
ABC. 

PROP.  IIL 


If  the  pole  of  a  great  circle  he  the  same  with  the  intersection  of  other  two  great 
circles  :  the  arc  of  the  first  mentioned  circle  intercepted  between  the  other 
two,  is  the  measure  of  the  spherical  angle  which  the  same  two  circles  make 
with  one  another. 

Let  the  great  circles  BA,  CA  on  the  superficies 
of  a  sphere,  of  which  the  centre  is  D,  intersect  one 
another  in  A,  and  let  BC  be  an  arc  of  another  great 
circle,  of  which  the  pole  is  A  ;  BC  is  the  measure 
of  the  spherical  angle  BAC. 

Join  AD,  DB,  DC  ;  since  A  is  the  pole  of  BC, 
AB,  AC  are  quadrants  (2.),  and  the  angles  ADB, 
ADC  are  right  angles  :  therefore  (4.  def.  2.  Sup.), 
the  angle  CDB  is  the  inclination  of  the  planes  of 

31 


242 


SPHERICAL  TRIGONOMETRY. 


the  circles  AB,  AC,  and  is  (def.  3.)  equal  to  the  spherical  angle  BAG ; 
but  the  arc  BC  measures  the  angle  BDC,  therefore  it  also  measures  the 
spherical  angle  BAC* 

CoR.  If  two  arcs  of  great  circles,  AB  and  AC,  which  intersect  one  an- 
other in  A,  be  each  of  them  quadrants,  A  will  be  the  pole  of  the  great  cir- 
cle which  passes  through  E  and  C  the  extremities  of  those  arcs.  For 
since  the  arcs  AB  and  AC  are  quadrants,  the  angles  ADB,  ADC  are  right 
angles,  and  AD  is  therefore  perpendicular  to  the  plane  BDC,  that  is,  to  the 
plane  of  the  great  circle  which  passes  through  B  and  C.  The  point  A  is 
therefore  (1.  Cor.  2.)  the  pole  of  the  great  circle  which  passes  through  B 
and  C. 

PROP.  IV. 

If  the  planes  of  two  great  circles  of  a  sphere  he  at  right  angles  to  one  another ^ 
the  circumference  of  each  of  the  circles  passes  through  the  poles  of  the 
other  ;  and  if  the  circumference  of  one  great  circle  pass  through  the  poles 
of  another,  the  planes  of  these  circles  are  at  right  angles. 

Let  ACBD,  AEBF  be  two  great  circles,  the  planes  of  which  are  right 
angles  to  one  another,  the  poles  of  the  circle  AEBF  are  in  the  circumference 
ACBD,  and  the  poles  of  the  circle  ACBD  in  the  circumference  AEBF. 

From  G  the  centre  of  the  sphere,  draw  GC  in  the  plane  ACBD  perpen- 
dicular to  AB.  Then  because  GC  in  the  plane  ACBD,  at  right  angles 
to  the  plane  AEBF,  is  at  right  angles 
to  the  common  section  of  the  two 
planes,  it  is  (Def.  2.  2.  Sup.)  also  at 
right  angles  to  the  plane  AEBF,  and 
therefore  (1.  Cor.  2.)  C  is  the  pole  of 
the  circle  AEBF  ;  and  if  CG  be  pro- 
duced in  D,  D  is  the  other  pole  of  the 
circle  AEBF. 

In  the  same  manner,  by  drawing 
GE  in  the  plane  AEBF,  perpendicu- 
lar to  AB,  and  producing  it  to  F,  it  has 
shewn  that  E  and  F  are  the  poles  of 
the  circle  ACBD.  Therefore,  the 
poles  of  each  of  these  circles  are  in 
the  circumference  of  the  other. 

Again,  If  C  be  one  of  the  poles  of  the  circle  AEBF,  the  great  circle 
ACBD  which  passes  through  C,  is  at  right  angles  to  the  circle  AEBF. 
For,  CG  being  drawn  from  the  pole  to  the  centre  of  the  circle  AEBF,  is 
at  right  angles  (1.  Cor.  2.)  to  the  plane  of  that  circle  ;  and  therefore,  every 
plane  passing  through  CG  (17.  2.  Sup.)  is  at  right  angles  to  the  plane 
AEBF  ;  now,  the  plane  ACBD  passes  through  CG. 

Cor.  1.     If  of  two  great  circles,  the  first  passes  through  the  poles  of  the- 


♦  When  In  any  reference  no  mention  is  made  of  a  Book,  or  of  the  Plane  Trigonometry,  the 
Spherical  Trigonometry  is  meant. 


SPHERICAL  TRIGONOMETRY.  243 

second,  the  second  also  passes  through  the  poles  of  the  first.  For,  if  the 
first  passes  through  the  poles  of  the  second,  the  plane  of  the  first  must  be 
at  right  angles  to  the  plane  of  the  second,  by  the  second  part  of  this  propo- 
sition ;  and  therefore,  by  the  first  part  of  it,  the  circumference  of  each 
passes  through  the  poles  of  the  other. 

Cor.  2.  All  greater  circles  that  have  a  common  diameter  have  their 
poles  in  the  circumference  of  a  circle,  the  plane  of  which  is  perpendicular 
to  that  diameter. 


PROP.  V. 

In  isosceles  spherical  triangles  the  angles  at  the  base  are  equal. 

Let  ABC  be  a  spherical  triangle,  having  the  side  AB  equal  to  the  side 
AC  ;  the  spherical  angles  ABC  and  ACB  are  equal. 

Let  C  be  the  centre  of  the  sphere  ;  join 
DB,  DC,  DA,  and  from  A  on  the  straight 
lines  DB,  DC,  draw  the  perpendiculars  AE, 
AF ;  and  from  the  points  E  and  F  draw  in 
the  plane  DBC  the  straight  lines  EG,  FG 
perpendicular  to  DB  and  DC,  meeting  one 
another  in  G :  Join  AG. 

Because  DE  is  at  right  angles  to  each  of 
the  straight  lines  AE,  EG,  it  is  at  right  angles 
to  the  plane  AEG,  which  passes  through 
AE,  EG  (4.  2.  Sup.) ;  and  therefore,  every 
plane  that  passes  through  DE  is  at  right  angles  to  the  plane  AEG  (17.  2. 
Sup.) ;  wherefore,  the  plane  DBC  is  at  right  angles  to  the  plane  AEG. 
For  the  same  reason,  the  plane  DBC  is  at  right  angles  to  the  plane  AFG, 
and  therefore  AG,  the  common  section  of  the  planes  AFG,  AEG  is  at 
right  angles  (18.  2.  Sup.)  to  the  plane  DBC,  and  the  angles  AGE,  AGF 
are  consequently  right  angles. 

But  since  the  arc  AB  is  equal  to  the  arc  AC,  the  angle  ADB  is  equal 
to  the  angle  ADC.  Therefore  the  triangles  ADE,  ADF,  have  the  angles 
EDA,  FDA,  equal,  as  also  the  angles  AED,  AFD,  which  are  right  an- 
gles ;  and  they  have  the  side  AD  common,  therefore  the  other  sides  are 
equal,  viz.  AE  to  AF(26.  1.),  and  DE  to  DF.  Again,  because  the  angles 
AGE,  AGF  are  right  angles,  the  squares  on  AG  and  GE  are  equal  to  the 
square  of  AE  ;  and  the  squares  of  AG  and  GF  to  the  square  of  KF.  But 
the  squares  of  AE  and  AF  are  equal,  therefore  the  squares  of  AG  and  GE 
are  equal  to  the  squares  of  AG  and  GF,  and  taking  away  the  common 
square  of  AG,  the  remaining  squares  of  GE  and  GF  are  equal,  and  GE  is 
therefore  equal  to  GF.  Wherefore,  in  the  triangles  AFG,  AEG,  the  side 
GF  is  equal  to  the  side  GE,  and  AF  has  been  proved  to  be  equal  to  AE, 
and  the  base  AG  is  common  ;  therefore,  the  angle  AFG  is  equal  to  the 
angle  AEG  (8.  1.).  But  the  angle  AFG  is  the  angle  which  the  plane 
ADC  makes  with  the  plane  DBC  (4.  def.  2.  Sup.),  because  FA  and  FG, 
which  are  drawn  in  these  planes,  are  at  right  angles  to  DF,  the  common 
section  of  the  planes.     The  angle  AFG  (3.  def.)  is  therefore  equal  to  the 


244  SPHERICAL  TRIGONOMETRY. 

spherical  angle  ACB  ;  and,  for  the  same  reason,  the  angle  AEG  is  equal 
to  the  spherical  angle  ABC.  But  the  angles  AFG,  AEG  are  equal. 
Therefore  the  spherical  angles  ACB,  ABC  are  also  equal. 

PROP.  VI. 

If  the  angles  at  the  base  of  a  spherical  triangle  he  equals  the  triangle  is  isosceles. 

Let  ABC  be  a  spherical  triangle  having  the  angles  ABC,  ACB  equal 
to  one  another  ;  the  sides  AC  and  AB  are  also  equal. 

Let  D  be  the  centre  of  the  sphere  ;  join  DB,  DC,  DA,  and  from  A  on 
the  straight  lines  DB,  DC,  draw  the  perpendiculars  AE,  AF  ;  and  from 
the  points  E  and  F,  draw  in  the  plane  DBC  a 

the  straight  lines  EG,  FG  perpendicular  to 
DB  and  DC,  meeting  one  another  in  G ; 
join  AG. 

Then,  it  may  be  proved,  as  was  done  in         j     \\\  \  ^C 

the  last  proposition,  that  AG  is  at  right  an- 
gles to  the  plane  BCD,  and  that  therefore 
the  angles  AGF,  AGE  are  right  angles,  and 
also  that  the  angles  AFG,  AEG  are  equal 

to  the  angles  which  the  planes  DAC,  DAB        ^ 

make  with  the  plane  DBC.     But  because      ^  ^ 

the  spherical  angles  ACB,  ABC  are  equal,  the  angles  which  the  planes 
DAC,  DAB  make  with  the  plane  DBC  are  equal  (3.  def.),  and  therefore 
the  angles  AFG,  AEG  are  also  equal.  The  triangles  AGE,  AGF  have 
therefore  two  angles  of  the  one  equal  to  two  angles  of  the  other,  and  they 
have  also  the  side  AG  common,  wherefore  they  are  equal,  and  the  side  AF 
is  equal  to  the  side  AE. 

Again,  because  the  triangles  ADF,  ADE  are  right  angled  at  F  and  E, 
the  squares  of  DF  and  FA  are  equal  to  the  square  of  DA,  that  is,  to  the 
squares  of  DE  and  DA  ;  now,  the  square  of  AF  is  equal  to  the  square  of 
AE,  therefore  the  square  of  DF  is  equal  to  the  square  of  DE,  and  the  side 
DF  to  the  side  DE.  Therefore,  in  the  triangles  DAF,  DAE,  because  DF 
is  equal  to  DE  and  DA  common,  and  also  AF  equal  to  AE,  the  angle 
ADF  is  equal  to  the  angle  ADE ;  therefore  also  the  arcs  AC  and  AB, 
which  are  the  measures  of  the  angles  ADF,  and  ADE,  are  equal  to  one 
another  ;  and  the  triangle  ABC  is  isosceles. 


PROP.  VIL 

Any  two  sides  of  a  spherical  triangle  are  greater  than  the  third, 

liCt  ABC  be  a  spherical  triangle,  any  two  sides  AB,  BC  are  greater  than 
the  third  side  AC. 


SPHERICAL  TRIGONOMETRY.  ^  245 

Let  D  be  the  centre  of  the  sphere ; 
join  DA,  DB,  DC. 

The  solid  angle  at  D  is  contained  by- 
three  plane  angles  ADB,  ADC,  BDC  ; 
any  two  of  which,  ADB,  BDC  are 
greater  (20.  2.  Sup.)  than  the  third 
ADC  ;  and  therefore  any  two  of  the 
arcs  AB,  AC,  BC,  which  measure 
these  angles,  as  AB  and  BC  must  also 
be  greater  than  the  third  AC. 

PROP.  VIIL 

The  three  sides  of  a  spherical  triangle  are  less  than  the  circumference  of  a 

great  circle. 

Let  ABC  be  a  spherical  triangle  as  before,  the  three  sides  AB,  BC,  AC 
are  less  than  the  circumference  of  a  great  circle. 

Let  D  be  the  centre  of  the  sphere  :  The  solid  angle  at  D  is  contained 
by  three  plane  angles  BDA,  BDC,  ADC,  which  together  are  less  than 
four  right  angles  (21.2.  Sup.)  therefore  the  sides  AB,  BC,  AC,  which  are 
the  measures  of  these  angles,  are  together  less  than  four  quadrants  describ- 
ed with  the  radius  AD,  that  is,  than  the  circumference  of  a  great  circle. 

PROP.  IX. 

In  a  spherical  triangle  the  greater  angle  is  opposite  to  the  greater  side  ;  and 

conversely. 

Let  ABC  be  a  spherical  triangle,  the  greater  angle  A  is  opposed  to  the 
greater  side  BC. 

Let  the  angle  BAD  be  made  equal 
to  the  angle  B,  and  then  BD,  DA  will 
be  equal  (6.),  and  therefore  AD,  DC 
are  equal  to  BC ;  but  AD,  DC  are 
greater  than  AC  (7.),  therefore  BC  is 
greater  than  AC,  that  is,  the  greater 
angle  A  is  opposite  to  the  greater  side 
B  C.  The  converse  is  demonstrated  as 
Prop.  19.  1.  Elem. 

PROP.  X. 

According  as  the  sum  of  two  of  the  sides  of  a  spherical  triangle,  is  greater  than 
a  semicircle,  equal  to  it,  or  less,  each  of  the  interior  angles  at  the  base  is  greater 
than  the  exterior  and  opposite  angle  at  the  base,  equal  to  it,  or  less  ;  and  also 
the  sum  of  the  two  interior  angles  at  the  base  greater  than  two  right  angles, 
equal  to  two  right  angles,  or  less  than  two  right  angles. 

Let  ABC  be  a  spherical  triangle,  of  which  the  sides  are  AB  and  EC ; 


■^^x> 


246  SPHERICAL  TRIGONOMETRY. 

produce  any  of  the  twosides  as  AB,  and  the  base  AC,  till  they  meet  aeaiu 
m  p  ;  then,  the  arc  ABD  is  a  semicircle,  and  the  spherical  angles  at  A 
and  D  are  equal,  because  each  of  them  is  the  inclination  of  the  circle  ABD 
to  the  circle  ACD. 

1.  If  AB,  BC  be    equal  to  a  -n 

semicircle,  that  is,  to  AD,  BC  will  ^^"C*"  "^''^"x..^^ 

be  equal  to  BD,  and  therefore  (5.)  ^"^^     ^  ^^ 

the  angle  D,  or  the  angle  A,  will 
be  equal  to  the  angle  BCD,  that 
is,  the  interior  angle  at  the  base 
equal  to  the  exterior  and  oppo- 
site. 

2.  If  AB,  BC  together  be  greater  than  a  semicircle,  that  is,  greater  than 
ABD,  BC  will  be  greater  than  BD  ;  and  therefore  (9.),  the  angle  D,  that 
is,  the  angle  A,  is  greater  than  the  ang-le  BCD. 

3.  In  the  same  manner  it  is  shewn,  if  AB,  BC  together  be  less  than  a 
semicircle,  that  the  angle  A  is  less  than  the  angle  BCD. 

Now,  since  the  angles  BCD,  BCA  are  equal  to  two  right  angles,  if  the 
angle  A  be  greater  than  BCD,  A  and  ACB  together  will  be  greater  than 
two  right  angles.  If  A  be  equal  to  BCD,  A  and  ACB  together,  will  be 
equal  to  two  right  angles  ;  and  if  A  be  less  than  BCD,  A  and  ACB  will 
be  less  than  two  right  angles. 

PROP.  XI. 

If  the  angular  points  of  a  spherical  triangle  he  made  the  polos  of  three  great 
circles,  these  three  circles  by  their  intersections  will  form  a  triangle,  which 
is  said  to  be  supplemental  to  the  former ;  and  the  two  triangles  are  such, 
that  the  sides  of  the  one  are  the  supplements  of  the  arcs  which  measure  the 
angles  of  the  other. 

Let  ABC  be  a  spherical  triangle  ;  and  from  the  points  A,  B,  and  C  as 
poles,  let  the  great  circles  FE,  ED,  DF  be  described,  intersecting  one  an- 
other in  F,  D  and  E  ;  the  sides  of  the  triangle  FED  are  the  supplement  of 
the  measures  of  the  angles  A,  B,  C,  viz.  FE  of  the  angle  BAC,  DE  of  the 
angle  ABC,  and  DF  of  the  angle  ACB  :  And  again,  AC  is  the  supplement 
of  the  angle  DFE,  AB  of  the  angle  FED,  and  BC  of  the  angle  EDF. 

Let  AB  produced  meet  DE,  EF  in  G,  M  ; 
let  AC  meet  FD,  FE  in  K,  L ;  and  let  BC 
meet  FD,  DE  in  N,  H. 

Since  A  is  the  pole  of  FE,  and  the  circle 
AC  passes  through  A,  EF  will  pass  through 
the  pole  of  AC  (1.  Cor.  4.)  and  since  AC 
passes  through  C,  the  pole  of  FD,  FD  will 
pass  through  the  pole  of  AC  ;  therefore  the 
pole  of  AC  is  in  the  point  F,  in  which  the 
arcs  DF,  EF  intersect  each  other.  In  the 
same  manner,  D  is  the  pole  of  BC,  and  E 
the  pole  of  AB. 

And  since  F,  E  are  the  poles  of  AL,  AM,  the  arcs  FL  and  EM  (2.)  are 


SPHERICAL  TRIGONOMETRY.  247 

quadrants,  and  FL,  EM  together,  that  is,  FE  and  ML  together,  are  equal 
to  a  semicircle.  But  since  A  is  the  pole  of  ML,  ML  is  the  measure  of  the 
angle  BAG  (3.),  consequently  FE  is  the  supplement  of  the  measure  of  the 
angle  BAG.  In  the  same  manner,  ED,  DF  are  the  supplements  of  the 
measures  of  the  angles  ABG,  BGA. 

Since  likewise  CN,  BH  are  quadrants,  CN  and  BH  together,  that  is, 
NH  and  BG  together,  are  equal  to  a  semicircle  ;  and  since  D  is  the  pole  of 
NH,  NH  is  the  measure  of  the  angle  FDE,  therefore  the  measure  of  the 
angle  FDE  is  the  supplement  of  the  side  BG.  In  the  same  manner,  it  is 
shewn  that  the  measures  of  the  angles  DEF,  EFD  are  the  supplements 
of  the  sides  AB,  AG  in  the  triangle  ABG. 

PROP.  XII. 

The  three  angles  of  a  spherical  triangle  are  greater  than  two,  and  less  than  stXf 

right  angles. 

The  measure  of  the  angles  A,  B,  G,  in  the  triangle  ABG,  together  with 
the  three  sides  of  the  supplemental  triangle  DEF,  are  (11.)  equal  to  three 
semicircles ;  but  the  three  sides  of  the  triangle  FDE,  are  (8.)  less  than  two 
semicircles  ;  therefore  the  measures  of  the  angles  A,  B,  G,  are  greater  than 
a  semicircle  ;  and  hence  the  angles  A,  B,  G  are  greater  than  two  right 
angles. 

And  because  the  interior  angles  of  any  triangle,  together  with  the  exte- 
rior, are  equal  to  six  right  angles,  the  interior  alone  are  less  than  six  right 
angles. 

PROP.  XIII. 

If  to  the  circumference  of  a  great  circle,  from  a  point  in  the  surface  of  the  sphere^ 
which  is  not  the  pole  of  that  circle,  arcs  of  great  circles  he  drawn ;  the  greatest 
of  these  arcs  is  that  which  passes  through  the  pole  of  the  first-mentioned  cir- 
cle, and  the  supplement  of  it  is  the  least ;  and  of  the  other  arcs,  that  which  is 
nearer  to  the  greatest  is  greater  than  that  which  is  more  remote. 

Let  ADB  be  the  circumference  of  a  great  circle,  of  which  the  pole  is  H, 
and  let  G  be  any  other  point ;  through  G  and  H  let  the  semicircle  AGB  be 
drawn  meeting  the  circle  ADB  in  A  and  B  ;  and  let  the  arcs  GD,  CE,  GF 
also  be  described.  From  G  draw  GG  perpendicular  to  AB,  and  then,  be- 
cause the  circle  AHGB  which  passes 
through  H,  the  pole  of  the  circle  ADB, 
is  at  right  angles  to  ADB,  GG  is  per- 
pendicular to  the  plane  ADB.  Join 
GD,  GE,  GF,  CA,  GD,  GE,  GF,  GB. 

Because  AB  is  the  diameter  of  the 
circle  ADB,  and  G  a  point  in  it,  which 
is  not  the  centre,  (for  the  centre  is  in 
the  point  where  the  perpendicular  from 
H  meets  AB),  therefore  AG,  the  part 
of  the  diameter  in  which  the  centre  is, 


248  SPHERICAL  TRIGONOMETRY. 

is  the  greatest  (7.  3.),  and  GB  the  least  of. all  the  straight  lines  that  can  be 
drawn  from  G  to  the  circumference  ;  and  GD,  which  is  nearer  to  AB,  is 
greater  than  GE,  which  is  more  remote.  But  the  triangles  CGA,  CGD 
are  right  angled  at  G,  and  therefore  AC2=AG2+GC2,  and  002=002+ 
GC2;  but  AG2+GC2  7DG2+G02;  because  AG 7 DO;  therefore  AC2 
7DC2,  and  AC  7  DC.  And  because  the  chord  AC  is  greater  than  the 
chord  DC,  the  arc  AC  is  greater  than  the  arc  DC.  In  the  same  manner, 
since  GD  is  greater  than  GE,  and  GE  than  GF,  it  is  shewn  that  CD  is 
greater  than  CE,  and  CE  than  OF.  Wherefore  also  the  arc  CD  is  greater 
than  the  arc  CE,  and  the  arc  GE  greater  than  the  arc  OF,  and  CF  than 
CB,  that  is,  of  all  the  arcs  of  greater  circles  drawn  from  0  to  the  circum- 
ference of  the  circle  ADB,  AC  which  passes  through  the  pole  H,  is  the 
greatest,  and  CB  its  supplement  is  the  least ;  and  of  the  others,  that  which 
is  nearer  to  AC  the  greatest,  is  greater  than  that  which  is  more  remote. 

PROP.  XIV. 

In  a  right  angled  spherical  triangle^  the  sides  containing  the  right  angle  are  oj 
the  same  affection  with  the  angles  opposite  to  them^  that  is,  if  the  sides  bd 
greater  or  less  than  quadrants,  the  opposite  angles  will  be  greater  or  less  than 
right  angles,  and  conversely. 

Let  ABC  be  a  spherical  triangle,  right  angled  at  A,  any  side  AB  will 
be  of  the  same  affection  with  the  opposite  angle  ACB. 

Produce  the  arcs  AC,  AB,  till  they  meet  again  in  D,  and  bisect  AD  in 
E.  Then  ACD,  ABD  are  semicircles,  and  AE  an  arc  of  90°.  Also,  be- 
cause CAB  is  by  hypothesis  a  right  angle,  the  plane  of  the  circle  ABD  is 
perpendicular  to  the  plane  of  the 

circle  ACD,  so  that  the  pole  of  Q. 

ACD  is   in  ABD,  (1.  Cor.  4.),  ^ 

and  is  therefore  the  point  E.    Let  ' 

EC  be  an  arc  of  a  great  circle 
passing  through  E  and  C. 

Then  because  E  is  the  pole  of 
the  circle  ACD,  EC  is  a  (2.) 
quadrant,  and  the  plane  of  the 
circle  EC  (4.)  is  at  right  angles 
to  the  plane  of  the  circle  ACD, 
that  is,  the  spherical  angle  ACE 
is  a  right  angle  ;  and  therefore, 
when  AB  is  less  than  AE,  the 
angle  ACB,  being  less  than 
ACE,  is  less  than  a  right  angle. 
But  when  AB  is  greater  than 
AE,  the  angle  ACB  is  gi'eater 
than  ACE,  or  than  a  right  an- 
gle. In  the  same  way  may  the 
converse  be  demonstrated. 


SPHERICAL  TRIGONOMETRY.  249 


PROP.  XV. 

If  the  two  sides  of  a  right  angled  spherical  triangle  about  the  right  angle  he  of 
the  same  affection  f  the  hypotenuse  will  be  less  than  a  quadrant ;  and  if  they  be 
of  different  affection,  the  hypotenuse  will  be  greater  than  a  quadrant. 

Let  ABC  be  a  right  angled  spherical  triangle  ;  according  as  the  two 
sides  AB,  AC  are  of  the  same  or  of  different  affection,  the  hypotenuse  BC 
will  be  less,  or  greater  than  a  quadrant. 

The  construction  of  the  last  proposition  remaining,  bisect  the  semicircle 
ACD  in  G,  then  AG  will  be  an  arc  of  90°,  and  G  will  be  the  pole  of  the 
circle  ABD. 

1.  Let  AB,  AC  be  each  less  than  90°.  Then,  because  C  is  a  point  on 
the  surface  of  the  sphere,  which  is  not  the  pole  of  the  circle  ABD,  the  arc 
CGD,  which  passes  through  G  the  pole  of  ABD  is  greater  than  CE  (13.), 
and  CE  greater  than  CB.  But  CE  is  a  quadrant,  as  was  before  shewn, 
therefore  CB  is  less  than  a  quadrant.  Thus  also  it  is  proved  of  the  right 
angled  triangle  CDB,  (right  angled  atD),  in  which  each  of  the  sides  CD, 
DB  is  greater  than  a  quadrant,  that  the  hypotenuse  BC  is  less  than  a 
quadrant. 

2.  Let  AC  be  less,  and  AB  greater  than  90°.  Then  because  CB  falls 
between  CGD  and  CE,  it  is  greater  (12.)  than  CE,  that  is,  than  a  quad- 
rant. 

CoR,  1.  Hence  conversely,  if  the  hypotenuse  of  a  right  angled  triangle 
be  greater  or  less  than  a  quadrant,  the  sides  will  be  of  different  or  the  same 
affection. 

CoR.  2.  Since  (14.)  the  oblique  angles  of  a  right  angled  spherical  trian- 
gle have  the  same  affection  with  the  opposite  sides,  therefore,  according  as 
the  hypotenuse  is  greater  or  less  than  a  quadrant,  the  oblique  angles  will 
be  different,  or  of  the  same  affection. 

Cor.  3.  Because  the  sides  are  of  the  same  affection  with  the  opposite 
angles,  therefore  when  an  angle  and  the  side  adjacent  are  of  the  same  affec- 
tion, the  hypotenuse  is  less  than  a  quadrant ;  and  conversely. 

PROP.  XVI. 

In  any  spherical  triangle,  if  the  perpendicular  upon  the  base  from  the  opposite 
angle  fall  within  the  triangle,  the  angles  at  the  base  are  of  the  same  affection; 
and  if  the  perpendicular  fall  without  the  triangle,  the  angles  at  the  base  are  of 
different  affection. 

Let  ABC  be  a  spherical  triangle,  and  let  the  arc  CD  be  drawn  from  C 
perpendicular  to  the  base  AB. 

1.  Let  CD  fall  within  the  triangle  ;  then,  since  ADC,  BDC  are  right 
angled  spherical  triangles,  the  angles  A,  B  must  each  be  of  the  same  affec- 
tion with  CD  (14.). 

33 


250 


SPHERICAL  TRIGONOMETRY. 
C 


2.  Let  CD  fall  without  the  triangle  ;  then  (14.)  the  angle  B  is  of  the 
same  affection  with  CD  ;  and  the  angle  CAD  is  of  the  same  affection  with 
CD ;  therefore  the  angle  CAD  and  B  are  of  the  same  affection,  and  the 
angle  CAB  and  B  are  therefore  of  different  affections. 

CoR.  Hence,  if  the  angles  A  and  B  be  of  the  same  affection,  the  per- 
pendicular will  fall  within  the  base ;  for  if  it  did  not,  A  and  B  would  be  of 
different  affection.  And  if  the  angles  A  and  B  be  of  different  affection, 
the  perpendicular  will  fall  without  the  triangle  ;  for,  if  it  did  not,  the  angles 
A  and  B  would  be  of  the  same  affection,  contrary  to  the  supposition. 


PROP.  XVII. 

If  to  the  base  of  a  spherical  triangle  a  perpendicular  he  drawn  from  the  opposite 
angle,  which  either  falls  within  the  triangle,  or  is  the  nearest  of  the  two  that 
fall  without;  the  least  of  the  segments  of  the  base  is  adjacent  to  the  least  of 
the  sides  of  the  triangle,  or  to  the  greatest,  according  as  the  sum  of  the  sides 
is  less  or  greater  than  a  semicircle. 

Let  ABEF  be  a  great  circle  of  a  sphere,  H  its  pole,  and  GHD  any  cir- 
cle passing  through  H,  which  therefore  is  perpendicular  to  the  circle 
ABEF.  Let  A  and  B  be  two  points  in  the  circle  ABEF,  on  opposite 
sides  of  the  point  D,  and  let  D  be  nearer 
to  A  than  to  B,  and  let  C  be  any  point 
in  the  circle  GHD  between  H  and  D. 
Through  the  points  A  and  C,  B  and  C, 
let  the  arcs  AC  and  BC  be  drawn,  and 
let  them  be  produced  till  they  meet  the 
circle  ABEF  in  the  points  E  and  F, 
then  the  arcs  ACE,  BCF  are  semicir- 
cles. Also  ACB,  ACF,  CFE,  ECB, 
are  four  spherical  triangles  continued 
by  arcs  of  the  same  circles,  and  having 
the  same  perpendiculars  CD  and  CG. 

I.  Now  because  CE  is  nearer  to  the  arc  CHG  than  CB  is,  CE  is  greater 
than  CA,  and  therefore  CE  and  CA  are  greater  than  CB  and  CA,  where- 
fore CB  and  CA  are  less  than  a  semicircle ;  but  because  AD  is  by  sup- 
position less  than  DB,  AC  is  also  less  than  CB  (13.),  and  therefore  in  this 
case,  viz.  when  the  perpendicular  falls  within  the  triangle,  and  when  the 


SPHERICAL  TRIGONOMETRY.  251 

sura  of  the  sides  is  less  than  a  semicircle,  the  last  segment  is  adjacent  to  the 
least  side. 

2.  Again,  in  the  triangle  FCA  the  two  sides  FC  and  CA  are  less  than 
a  semicircle  ;  for  since  AC  is  less  than  CB,  AC  and  CF  are  less  than  BC 
and  CF.  Also,  AC  is  less  than  CF,  because  it  is  more  remote  from  CHG 
than  CF  is ;  therefore  in  this  case  also,  viz.  when  the  perpendicular  falls 
without  the  triangle,  and  when  the  sum  of  the  sides  is  less  than  a  semicir- 
cle, the  least  segment  of  the  base  AD  is  adjacent  to  the  least  side. 

3.  But  in  the  triangle  FCE  the  two  sides  FC  and  CE  are  greater  than 
a  semicircle ;  for,  since  FC  is  greater  than  CA,  FC  and  CE  are  greater 
than  AC  and  CE.  And  because  AC  is  less  than  CB,  EC  is  greater  than 
CF,  and  EC  is  therefore  nearer  to  the  perpendicular  CHG  than  CF  is, 
wherefore  EG  is  the  least  segment  of  the  base,  and  is  adjacent  to  the 
greater  side. 

4.  In  the  triangle  ECB  the  two  sides  EC,  CB  are  greater  than  a  semi- 
circle ;  for,  since  by  supposition  CB  is  greater  than  CA,  EC  and  CB  are 
greater  than  EC  and  CA.  Also,  EC  is  greater  than  CB,  wherefore  in 
this  case,  also,  the  least  segment  of  the  base  EG  is  adjacent  to  the  greatest 
side  of  the  triangle.  Therefore,  when  the  sura  of  the  sides  is  greater  than 
a  seraicircle,  the  least  segment  of  the  base  is  adjacent  to  the  greatest  side, 
whether  the  perpendicular  fall  within  or  without  the  triangle  :  and  it  has 
been  shewn,  that  when  the  sum  of  the  sides  is  less  than  a  semicircle,  the 
least  segment  of  the  base  is  adjacent  to  the  least  of  the  sides,  whether  the 
perpendicular  fall  within  or  without  the  triangle. 

PROP.  XVIII. 

In  right  angled  spherical  triangles,  the  sine  of  either  of  the  sides  ahout  the  right 
angle,  is  to  the  radius  of  the  sphere,  as  the  tangent  of  the  remaining  side  is 
to  the  tangent  of  the  angle  opposite  to  that  side. 

Let  ABC  be  a  triangle,  having  the  right  angle  at  A ;  and  let  AB  be 
either  of  the  sides,  the  sine  of  the  side  AB  will  be  to  the  radius,  as  the  tan- 
gent of  the  other  side  AC  to  the  tangent  of  the  angle  ABC,  opposite  to  AC. 
Let  D  be  the  centre  of  the  sphere ;  join  AD,  BD,  CD,  and  let  AF  be  drawn 
perpendicular  to  BD,  which  therefore  will  be  the  sine  of  the  arc  AB,  and 
from  the  point  F,  let  there  be  drawn  in  the  plane  BDC  the  straight  line 
FE  at  right  angles  to  BD,  meeting  DC  in 
E,  and  let  AE  be  joined.  Since  therefore 
the  straight  line  DE  is  at  right  angles  to 
both  FA  and  FE,  it  will  also  be  at  right 
angles  to  the  plane  AEF  (4.  2.  Sup.)  ; 
wherefore  the  plane  ABD,  which  passes 
through  DF,  is  perpendicular  to  the  plane 
AEF  (17.  2.  Sup.),  and  the  plane  AEF 
perpendicular  to  ABD  :  But  the  plane 
ACD  or  AED,  is  also  perpendicular  to 
the  same  ABD,  because  the  spherical  an- 
gle BAC  is  a  right  angle  .  Therefore  AE, 
the  common  section  of  the  planes  AED, 


252  SPHERICAL  TRIGONOMETRY. 

AEF,  is  at  right  angles  to  the  plane  ABD  (18.  2.  Sup.),  and  EAF,  EAD 
are  right  angles.  Therefore  AE  is  the  tangent  of  the  arc  AC  ;  and  in  the 
rectiUneal  triangle  AEF,  having  a  right  angle  at  A,  AF  is  to  the  radius  as 
AE  to  the  tangent  of  the  angle  AFE  (1.  PL  Tr.) ;  but  AF  is  the  sine  of 
the  arc  AB,  and  AE  the  tangent  of  the  arc  AC  ;  and  the  angle  AFE  is 
the  inclination  of  the  planes  CBD,  ABD  (4.  def.  2.  Sup.),  or  is  equal  to  the 
spherical  angle  ABC  :  Therefore  the  sine  of  the  arc  AB  is  to  the  radius  as 
the  tangent  of  the  arc  AC  to  the  tangent  of  the  opposite  angle  ABC. 

CoR.  Since  by  this  proposition,  sin.  AB  :  R  : :  tan.  AC  :  tan.  ABC  ; 
and  because  R  :  cot.  ABC  :  :  tan.  ABC  :  R  (1  Cor.  def.  9.  PI.  Tr.)  by 
equality,  sin.  AB  :  cot.  ABC  :  :  tan.  AC  :  R. 

PROP.  XIX. 

In  right  angled  spherical  triangles  the  sine  of  the  hypotenuse  is  to  the  radius  as 
the  sine  of  either  side  is  to  the  sine  of  the  angle  opposite  to  that  side. 

Let  the  triangle  ABC  be  right  angled  at  A,  and  let  AC  be  either  of  the 
sides ;  the  sine  of  the  hypotenuse  BC  will  be  to  the  radius  as  the  sine  of 
the  arc  AC  is  to  the  sine  of  the  angle  ABC. 

Let  D  be  the  centre  of  the  sphere,  and  let  CE  be  drawn  perpendicular 
to  DB,  which  will  therefore  be  the  sine  of  the  hypotenuse  BC  ;  and  from 
the  point  E  let  there  be  drawn  in  the 
plane  ABD  the  straight  line  EF  per- 
pendicular to  DB,  and  let  CF  be  joined ; 
then  CF  will  be  at  right  angles  to  the 
plane  ABD,  because  as  was  shewn  of 
EA  in  the  preceding  proposition,  it  is 
the  common  section  of  two  planes  DCF, 
ECF,  each  perpendicular  to  the  plane 
ADB.  Wherefore  CFD,  CFE  are  right 
angles,  and  CF  is  the  sine  of  the  arc  _^ 

AC  ;  and  in  the  triangle  CFE  having  ^ 

the  right  angle  CFE,  CE  is  to  the  radius,  as  CF  to  the  sine  of  the  angle 
CEF  (1.  PL  Tr.).  But,  since  CE,  FE  are  at  right  angles  to  DEB,  which 
is  the  common  section  of  the  planes  CBD,  ABD,  the  angle  CEF  is  equal 
to  the  inclination  of  these  planes  (4.  def.  2.  Sup.),  that  is,  to  the  spherical 
angle  ABC.  Therefore  the  sine  of  the  hypotenuse  CB,  is  to  the  radius,  as 
the  sine  of  the  side  AC  to  the  sine  of  the  opposite  angle  ABC. 

PROP.  XX. 

In  right  angled  spherical  triangles,  the  cosine  of  the  hypotenuse  is  to  the  radius 
as  the  cotangent  of  either  of  the  angles  is  to  the  tangent  of  the  remaining 
angle. 

Let  ABC  be  a  spherical  triangle,  having  a  right  angle  at  A,  the  cosine 
of  the  hypotenuse  BC  is  to  the  radius  as  the  cotangent  of  the  angle  ABC 
to  the  tangent  of  the  angle  ACB, 


SPHERICAL  TRIGONOMETRY. 


253 


Describe  the  circle  DE,  of  which  B  is  the  pole,  and  let  it  meet  AC  in 
F.  and  the  circle  BC  in  E  ;  and  since  the  circle  BD  pases  through  the 


pole  B,  of  the  circle  DF,  BF  must  pass  through  the  pole  of  BD  (4.).  And 
since  AC  is  perpendicular  to  BD,  the  plane  of  the  circle  AC  is  perpendi- 
cular to  the  plane  of  the  circle  BAD,  and  therefore  AC  must  also  (4.)  pass 
through  the  pole  of  BAD  ;  wherefore,  the  pole  of  the  circle  BAD  is  in  the 
point  F,  where  the  circles  AC,  DE,  intersect.  The  arcs  FA,  FD  are 
therefore  quadrants,  and  likewise  the  arcs  BD,  BE.  Therefore,  in  the  tri- 
angle CEF,  right  angled  at  the  point  E,  CE  is  the  complement  of  BC,  the 
hypotenuse  of  the  triangle  ABC  ;  EF  is  the  complement  of  the  arc  ED, 
the  measure  of  the  angle  ABC,  and  FC,  the  hypotenuse  of  the  triangle 
CEF,  is  the  complement  of  AC,  and  the  arc  AD,  which  is  the  measure  of 
the  angle  CFE,  is  the  complement  of  AB. 

But  (18.)  in  the  triangle  CEF,  sin.  CE  :  R  :  :  tan.  £F  :  tan.  ECF,  that 
is,  in  the  triangle  ACB,  cos.  BC  :  R  :  :  cot.  ABC  :  tan.  ACB. 

CoR.  Because  cos.  BC  :  R  : :  cot.  ABC  :  tan.  ACB,  and  (Cor.  1.  def.  9. 
PI.  Tr.)  cot.  ABC  :  R  :  :  R  :  tan.  ABC,  ex  aequo,  cot.  ACB  :  cos.  BC  :  :  R 
:  cot.  ABC, 


PROP.    XXI. 

In  right  angled  spherical  triangles^  the  cosine  of  an  angle  is  to  the  radius  as  the 
tangent  of  the  side  adjacent  to  that  angle  is  to  the  tangent  of  the  hypotenuse. 

The  same  construction  remaining ;  In  the  triangle  CEF,  sin.  FE  :  R  : : 
tan.CE  :  tan.  CFE(18.):  butsin.  EF=cos.  ABC  ;  tan.  CE=cot.BC,and 
tan.  CFE=cot.  AB,  therefore  cos.  ABC  :  R  :  :  cot.  BC  :  cot.  AB.  Now, 
because  (Cor.  1.  def.  9.  PL  Tr.)  cot.  BC  :  R  :  :  R  :  tan.  BC,  and  cot.  AB  : 
R  :  :  R  :  tan.  AB,  by  equality  inversely,  cot.  BC  :  cot.  AB  :  :  tan.  AB  : 
BC ;  therefore  (11.  5.)  cos.  ABC  :  R  :  :  tan.  AB  :  tan.  BC. 

Cor.  1.  From  the  demonstration  it  is  manifest,  that  the  tangents  of  any 
two  arcs  AB,  BC  are  reciprocally  proportional  to  their  cotangents. 


254  SPHERICAL  TRIGONOMETRY 


Cor.  2.  Because  cos.  ABC  :  R  :  :  tan.  AB  :  tan.  BC,  and  R  :  cos.  BC  : : 
tan.  BC  :  R,  by  equality,  cos.  ABC  :  cot.  BC  :  :  tan.  AB  :  R.  That  is,  the 
cosine  of  any  of  the  oblique  angles  is  to  the  cotangent  of  the  hypotenuse, 
as  the  tangent  of  the  side  adjacent  to  the  angle  is  to  the  radius. 

PROP.  XXII. 

In  right  angled  spherical  triangles,  the  cosine  of  either  of  the  sides  is  to  the  ro" 
diuSf  as  the  cosine  of  the  hypotenuse  is  to  the  cosine  of  the  other  side. 

The  same  construction  remaining :  In  the  triangle  CEF,  sin.  CF  :  R  :  : 
sin.  CE  :  sin.  CFE  (19.) ;  but  sin.  CF=cos.  CA,  sin.  CE=cos.  BC,  and 
sin.  CFE=cos.  AB ;  therefore  cos.  CA  :  R  :  :  cos.  BC  :  cos.  AB. 


PROP.  XXIII. 

In  right  angled  spherical  triangles,  the  cosine  of  either  of  the  sides  is  to  the  ra- 
dius, as  the  cosine  of  the  angle  opposite  to  that  side  is  to  the  sine  of  the  other 
angle. 

The  same  construction  remaining :  In  the  triangle  CEF,  sin.  CF  :  R  : : 
sin.  EF  :  sin.  ECF  (19.) ;  but  sin.  CF=cos.  CA,  sin.  EF=cos.  ABC,  and 
sin.  ECF=sin.  BCA :  therefore,  cos.  CA  :  R  :  :  cos.  ABC  :  sin.  BCA. 

PROP.  XXIV. 

In  spherical  triangles,  whether  right  angled  or  oblique  angled,  the  sines  of  the 
sides  are  proportional  to  the  sines  of  the  angles  opposite  to  them. 

First,  let  ABC  be  a  right  angled  triangle,  having  a  right  angle  at  A ; 
therefore  (19.),  the  sine  of  the  hypotenuse  BC  is  to  the  radius,  (or  the  sine 


SPHERICAL  TRIGONOMETRY. 


255 


B 


of  the  right  angle  at  A),  as  the  sine  of 
the  side  AC  to  the  sine  of  the  angle  B. 
And,  in  like  manner,  the  sine  of  BC  is 
to  the  sine  of  the  angle  A,  as  the  sine 
of  AB  to  the  sine  of  the  angle  C ; 
wherefore  (11.  5.)  the  sine  of  the  side 
AC  is  to  the  sine  of  the  angle  B,  as  the 
sine  of  A  B  to  the  sine  of  the  angle  C. 

Secondly,  Let  ABC  be  an  oblique  angled  triangle,  the  sine  of  any  of  the 
sides  BC  will  be  to  the  sine  of  any  of  the  other  two  AC,  as  the  sine  of  the 
angle  A  opposite  to  BC,  is  to  the  sine  of  the  angle  B  opposite  to  AC. 
Through  the  point  C,  let  there  be  drawn  an  arc  of  a  great  circle  CD  per- 
pendicular to  AB  ;  and  in  the  right  angled  triangle  BCD,  sin.  BC  :  R  : ; 


sin.  CD  :  sin.  B  (19.) ;  and  in  the  triangle  ADC,  sin.  AC  ;  R  :  :  sin.  CD  : 
sin.  A;  wherefore,  by  equality  inversely,  sin.  BC  :  sin.  AC  : :  sin.  A  :  sin. 
B.  In  the  same  manner,  it  may  be  proved  that  sin.  BC  :  sin.  AB  : :  sin. 
A  :  sin.  C,  &c. 

PROP.  XXV. 

In  oblique  angled  spherical  triangles^  a  perpendicular  arc  being  drawn  from 
any  of  the  angles  upon  the  opposite  side,  the  cosines  of  the  angles  at  the  base 
are  proportional  to  the  sines  of  the  segments  of  the  vertical  angle. 

Let  ABC  be  a  triangle,  and  the  arc  CD  perpendicular  to  the  base  BA  ; 
the  cosine  of  the  angle  B  will  be  to  the  cosine  of  the  angle  A,  as  the  sine 
of  the  angle  BCD  to  the  sine  of  the  angle  ACD. 

For  having  drawn  CD  perpendicular  to  AB,  in  the  right  angled  triangle 
BCD  (23.),  cos.  CD  :  R  : :  cos.  B  :  sin.  DCB ;  and  in  the  right  angled 
triangle  ACD,  cos.  CD  :  R  : :  cos.  A  :  sin.  ACD ;  therefore  (11.  5.)  cos. 
B  :  sin.  DCB  : :  cos.  A  :  sin.  ACD,  and  alternately,  cos.  B  :  cos.  A  : :  sin. 
BCD  :  sin.  ACD. 


PROP.  XXVI. 

The  same  things  remaining,  the  cosines  of  the  sides  BC,  CA,  are  proportional 
to  the  cosines  o/*BD,  DA,  the  segments  of  the  base. 

For  in  the  triangle  BCD  (22.),  cos.  BC  :  cos.  BD  :  :  cos.  DC  :  R,  and  in 


2S6  SPHERICAL  TRIGONOMETRY. 

the  triangle  ACD,  cos.  AC  :  cos.  AD  : :  cos.  DC  :  R ;  therefore  (11.  5.) 
COS.  BC  :  COS.  BD  :  :  cos.  AC  :  cos.  AD,  and  alternately,  cos.  BC  :  cos. 
AC  : :  cos.  BD  :  cos.  AD. 

PROP,  XXVII. 

The  same  construction  remaining^  the  sines  of  BD,  DA,  the  segments  of  the 
base  are  reciprocally  proportional  to  the  tangents  ofB  and  A,  the  angles 
at  the  base. 

In  the  triangle  BCD  (18.),  sin.  BD  :  R  ; :  tan.  DC  :  tan.  B  ;  and  in  the 
triangle  ACD,  sin.  AD  :  R  : :  tan.  DC  :  tan.  A ;  therefore,  by  equality  in- 
versely, sin.  BD  :  sin.  AD  ; ;  tan.  A  :  tan.  B. 


PROP.  XXVIII. 

The  same  construction  remaining^  the  cosines  of  the  segments  of  the  vertical 
angle  are  reciprocally  proportional  to  the  tangents  of  the  sides. 

Because  (21.),  cos.  BCD  :  R  : :  tan.  CD  :  tan.  BC,  and  also  cos.  ACD 
R  :  :  tan.  CD  :  tan.  AC,  by  equality  inversely,  cos.  BCD  ;  cos.  ACD  : : 
tan.  AC  :  tan.  BC. 

PROP.  XXIX. 

If  from  an  angle  of  a  spherical  triangle  there  be  dravm  a  perpendicular  to  the 
opposite  side,  or  base,  the  rectangle  contained  by  the  tangents  of  half  the 
sum,  and  of  half  the  difference  of  the  segments  of  the  base  is  equal  to  the 
rectangles  contained  by  the  tangents  of  half  the  sum,  and  of  half  the  diffe- 
rence of  the  two  sides  of  the  triangle. 

Let  ABC  be  a  spherical  triangle,  and  let  the  arc  CD  be  drawn  from  the 
angle  C  at  right  angles  to  the  base  AB,  tan.  ^  (m-fn)  Xtan.  J  (m— w)=J 
tan.  (a+5)xjtan.  (a—b). 

Let  BC=a,  AC=b  ;  BD=m,  AD=n.  Because  (26.)  cos.  a  :  cos.  b  : : 
cos.  m :  COS.  n(E.  5.),  cos.  a-\-b  :  cos.  a — cos.  b  : :  cos.  m+cos.  n  :  cos.  m — 
cos,  n.  But  (1.  Cor.  3.  PI.  Trig.),  cos.  a-fcos.  b  :  cos.  a—cos.  b  : :  cot.  J 
(a-\-b)  :  tan.  -J  [a — b),  and  also,  cos.  m+cos.  n  :  cos.  m — cos.  n  :  :  cot.  -J 
(w»4-n)  :  tan.  J  (m—n).  Therefore,  (11.  5.)  cot.  ^  (a-^b)  :  tan.  -^  (a—b) 
: ;  cot.  ^  (m-\-n)  :  tan.  J  [m — n).     And  because  rectangles  of  the  same  al- 


SPHERICAL  TRIGONOMETRY.  257 

titude  are  as  their  bases,  tan.  |  (a+^)Xcot.  ^{a-^h)  :  tan.  I  (<z4-^)Xtan. 
-J-  (a— i)  ::  tan.  J  {m-{-n)x  cot.  ^  (m-fn)  :  tan.  J  {mXn)-{-ian.  ^{m—n). 
Now  the  first  and  third  terms  of  this  proportion  are  equal,  being  each  equal 
to  the  square  of  the  radius  (1.  Cor.  PI.  Trig.),  therefore  the  remaining  two 
are  equal  (9.  5.),  or  tan.  ^{m-\-n)xtn,n.  -J  (m— n)=tan.  J  (a+6)X  tan.  ^ 
{a—b) ;  that  is,  tan.  i  (BD+ AD)  X  tan.  i  (BD--AD)=tan.  i  (BC+ AC) 
Xtan.  i(BC-AC). 

CoR.  1.  Because  the  sides  of  equal  rectangles  are  reciprocally  propor- 
tional, tan.  i  (BD+AD)  :  tan.  I  (BC+AC)  :  :  tan.  I  (BC  —  AC)  :  tan.  4 
(BD-AD). 

Cor.  2.  Since,  when  the  perpendicular  CD  falls  within  the  triangle, 
BD4-AD=AB,  the  base  ;  and  when  CD  falls  without  the  triangle  BD — 
AD=AB,  therefore,  in  the  first  case,  the  proportion  in  the  last  corollary 
becomes  tan. i(AB)  :  tan.  J  (BC+AC)  ::  tan.-^(BC— AC)  :  tan.i(BD— 
AD) ;  and  in  the  second  case,  it  becomes  by  inversion  and  alternation,  tan. 
J  (AB)  ;  tan.  J  (BC+AC)  : :  tan.  J  (BC-AC)  ;  tan.  ^  (BD+AD). 


SCHOLIUM. 

The  preceding  proposition,  which  is  very  useful  in  spherical  trigonome- 
try, may  be  easily  remembered  from  its  analogy  to  the  proposition  in  plane 
trigonometry,  that  the  rectangle  under  half  the  sum,  and  half  the  difference 
of  the  sides  of  a  plane  triangle,  is  equal  to  the  rectangle  under  half  the 
sum,  and  half  the  difference  of  the  segments  of  the  base.  See  (K.  6.),  also 
4th  Case  PI.  Tr.  We  are  indebted  to  Napier  for  this  and  the  two  follow- 
ing theorems,  which  are  so  well  adapted  to  calculation  by  Logarithms, 
that  they  must  be  considered  as  three  of  the  most  valuable  propositions  in 
Trigonometry. 

33 


258 


SPHERICAL  TRIGONOMETRY. 


PROP.  XXX. 

If  a  perpendicular  be  drawn  from  an  angle  of  a  spherical  triangle  to  the  oppo- 
site  side  or  base,  the  sine  of  the  sum  of  the  angles  at  the  base  is  to  the  sine 
of  their  difference  as  the  tangent  of  half  the  base  to  the  tangent  of  half  the 
difference  of  its  segments,  when  the  perpendicular  falls  within ;  but  as  the 
co-tangent  of  half  the  base  to  the  co-tangent  of  half  the  sum  of  the  segments, 
when  the  perpendicular  falls  without  the  triangle  :  And  the  sine  of  the  sum 
of  the  two  sides  is  to  the  sine  of  their  difference  as  the  co-tangent  of  half 
the  angle  contained  by  the  sides,  to  the  tangent  of  half  the  difference  of 
the  angles  which  the  perpendicular  makes  with  the  same  sides  when  it  falls 
within,  or  to  the  tangent  of  half  the  sum  of  these  angles,  when  it  falls  with- 
out the  triangle. 

If  ABC  be  a  spherical  triangle,  and  AD  a  perpendicular  to  the  base  BC, 
sin.  (C+B)  :  sin.  (C-B)  : :  tan.  i  BC  :  tan.  J  (BD-DC),  when  AD  falls 
within  the  triangle ;  but  sin.  (C+B)  :  sin.  (C— B)  :  :  cot.  J  BC  :  cot.  J 
(BD-f  DC),  when  AC  falls  without.     And  again, 

A 


B 


3> 


C 


sin.  (AB+AC)  :  sin.  (AB— AC)  : :  cot.  i  BAG  :  tan.  i  (BAD— CAD), 
when  AD  falls  within ;  but  when  AD  falls  without  the  triangle, 
sin.  (AB+AC)  :  sin.  (AB-AC)  :  :  cot.  i  BAC  :  tan.  J  (BAD+CAD). 
For  in  the  triangle  BAC  (27.),  tan.  B  :  tan.  C  :  :  sin.  CD  :  sin.  BD,  and 
therefore  (E.  5.),  tan.  C+tan.  B  :  tan.  C— tan.  B  : :  sin.  BD+sin.  CD  : 
sin.  BD— sin.  CD.  Now  (by  the  annexed  Lemma),  tan.  C+tan.  B  :  tan. 
C— tan.  B  : :  sin.  (C+B)  :  sin.  (C— B),  and  sin.  BD+sin.  CD  :  sin.  BD 
-sin.  CD  :  :  tan.  i  (BD+CD)  :  tan.  i  (BD— CD),  (3.  PI.  Trig.),  there- 
fore because  ratios  which  are  equal  to  the  same  ratio  are  equal  to  one 
another  (11.  5.),  sin.  (C+B)  :  sin.  (C-B)  : :  tan.  -J  (BD+CD)  :  tan.  } 
(BD— CD). 


SPHERICAL  TRIGONOMETRY.  259 

Now  when  AD  is  within  the  triangle,  BD-f  CD=BC,  and  therefore  sin. 
(C+B)  :  sin.  (C— B)  :  :  tan.  i  BC  :  tan.  i  (BD— CD).  And  again,  when 
AD  is  without  the  triangle,  BD— CD =BC,  and  therefore  sin.  (C+B) :  sin. 
(C— B)  : :  tan.  J  (BD+CD)  :  tan.  ^  BC,  or  because  the  tangents  of  any 
two  arcs  are  reciprocally  as  their  co-tangents,  in  (C-f-B)  :  sin.  (C— B)  : : 
cot.  J  BC  :  cot.  i  (BD+CD). 

The  second  part  of  the  proposition  is  next  to  be  demonstrated.   Because 
(28.)  tan.  AB  :  tan.  AC  : :  cos.  CAD  :  cos.  BAD,  tan.  AB+tan.  AC  :  tan. 
AB— tan,  AC  : :  cos.  CAD+cos.  BAD  :  cos.  CAD— cos.  BAD.      But 
(Lemma)  tan.  AB-ftan.  AC  :  tan.  AB— tan.  AC  : :  sin.  (AB+AC)  :  sin. 
(AB-AC),and  (1.  cor.  3.  PL  Trig.)  cos.  CAD+cos.  BAD  :  cos.  CID— 
cos.  BAD  : :  cot.i(BAD+CAD)  :  tan.  i  (BAD-CAD).     Therefore  (XL 
5.)  sin.  (AB+AC)  :  sin.  (AB— AC)  : :  cot.  l  (BAD+CAD) :  tan.  i  (BAD 
—CAD).     Now,  when  AD  is  within  the  triangle,  BAD-f-CAD=BAC, 
and  therefore  sin.  (AB-f  AC) :  sin.  (AB— AC) : :  cot.  I BAC  :  tan.  i  (BAD 
^CAD.) 
But  if  AD  be  without  the  triangle,  BAD— CAD=BAC,  and  therefore 
sin.  (AB-f  AC)  :  sin.  (AB-AC)  : ; 
cot.  i  (BAD+CAD)  :  tan.  i  BAC  ;  or  because 
cot.  I  (BAD+CAD)  :  tan.  i  BAC  : :  cot.  i  BAC  ; 
tan.  1  (BAD+CAD),  sin.  (AB+AC)  :  sin.  (AB— AC)  : :  cot.  4  BAC  : 
tan.  I  (BAD  +  CAD). 

LEMMA. 

The  sum  of  the  tangents  of  any  two  arcs,  is  to  the  difference  of  their  tangentSt 
as  the  sine  of  the  sum  of  the  arcs,  to  the  sine  of  their  difference. 

Let  A  and  B  be  two  arcs,  tan.  A+tan.  B  :  tan.  A— tan.  B  : :  sin.  (A+B) 

:  (A-B). 

For,  by  §6.  page  232,  sin.  A X  cos.  B+cos.  Ax  sin.  B=sin.  (A+B),  and 

.       r      J-  -J-       11  u  »         -n  sin.  A  ,  sin.  B       sin.  (A+B)      ,    ^ 

therefore  dividmgall  by  cos.  A  cos.  B, + — = ^ — - — '■—.  that 

cos.  A     cos.  B     cos.  A  X  cos.  B 

.    .  sin.  A     ^       ,   ^        .   ,         _       sin.  (A+B)         ^     , 

IS,  because r-=tan.  A,  tan.  A+tan.  B= ^ — ■ — -^'     In  the  same 

cos.  A  COS.  A  X  cos.  B 

manner  it  is  proved  that  tan.  A —tan.  B  =  ^^"'/    "~    ^— .    Therefore  tan.  A 

COS.  Ax  COS.  B 

+tan.  B  :  tan.  A— tan.  B  :  :  sin.  (A+B)  :  sin.  (A— B). 

.        PROP.  XXXI. 

The  sine  of  half  the  sum  of  any  two  angles  of  a  spherical  triangle  is  to  tJie 
sine  of  half  their  difference,  as  the  tangent  of  half  the  side  adjacent  to  these 
angles  is  to  the  tangent  of  half  the  difference  of  the  sides  opposite  to  them  ; 
and  the  cosine  of  half  the  sum  of  the  same  angles  is  to  the  cosine  of  half 
their  difference,  as  the  tangent  of  half  the  side  adjacent  to  them,  to  the  tan- 
gent of  half  the  sum  of  the  sides  opposite. 

LetC+B=2S,  C— B=2D,  the  base  BC=2B,  and  the  difference  of 


260  SPHERICAL  TRIGONOMETRY. 

the  segments  of  the  base,  or  BD— CD=2X.  Then,  because  (30.)  sin. 
(C+B)  :  sin.  (C-B)  :  :  tan.  J  BC  :  tan.  i  (BD— CD),  sin.  2S  :  sin.  2D 
: :  tan.  B  :  tan.  X.  Now,  sin.  2S=sin.  (S+S)=2  sin.  Sx  cos.  S,  (Sect. 
III.  cor.  PI.  Tr.).  In  the  same  manner,  sin.  2D=2  sin.  Dxcos.  D 
Therefore  sin.  S  X  cos.  S  ;  sin.  D  X  cos.  D  : :  tan.  B  :  tan.  X. 


Again,  in  the  spherical  triangle  ABC  it  has  been  proved,  that  sin,  0+ 
sin.  B  :  sin.  C — sin.  B  :  :  sin.  AB  +  sin.  AC  :  sin.  AB — sin.  AC,  and  since 
sin.  C+sin.  B=2  sin.  J  (C+B)+cos.  J  (C-B),  (Sect.  III.  7.  PI.  Tr.)= 
2  sin.  S  X  cos.  D  ;  and  sin.  C— sin.  B=2  cos.  J  (C+B)  X  sin.  ^  (C— B)= 
2  cos.  S  X  sin.  D.  Therefore  3  sin.  S  X  cos.  D  :  2  cos.  S  x  sin.  D  : :  sin. 
AB+sin.  AC  :  sin.  AB— sin.  AC.  But  (3.  PI.  Tr.)  sin.  AB+sin.  AC  : 
sin.  AB— sin.  AC  : :  tan.  i  (AB+AC)  :  tan.  J  (AB— AC)  : :  tan.  2  :  tan. 
J,  2  being  equal  to  J  (AB+AC)  and  -i/  to  J  (AB— AC).     Therefore  sin. 

Sxcos,  D  :  cos.  Sxsin.  D  :  :  tan.  S  :  tan.  J.     Since  then  : — ^^  = 

tan.  o 

sin.  D  X  COS.  D         .  tan.  J    cos.  S  X  sin.  D    ,  ,  •  ,  •  i     . 

—■ — ET t;  ;  and ==-: — r; 7:r,  by   multiplymg   equals   by 

sm.  Sxcos.  S'  tan.  ^     sm.  Sxcos.  D'     ^  tj    a     ^  j 

-     tan.  X     tan.  ^     (sin.  D)2xcos.  Sxcos.  D     (sin.  D)2 


tan.  B     tan.  2     (sin.  S)2xcos.  Sxcos.  D     (sin.  S) 

B„t  foQ  >  ^^"-  ^  (BD-DC)_tan.l(AB+AC)  tan.  X__tan.  2 

^"^  ^^^'^  tan.  J  (AB- AC)-       tan.  J  BC      '  ^^^* '''  i^iT^-tan.  B' 

,  ,        .       tan.  X     ta.n.2xt2in.J  ,       tan.  X     tan.  ^      tan.  .^ 

and  therefore, -= — ; -— — ,  as  also =r= ^=7: 5^- 

tan.  B  (tan.  B)^     '  tan.  B     tan.  2     (tan.  B)^ 

^     tan.  X     tan.  J     (sin.  D)2        ,  (tan.  ^V     (sin.  D)^         ,  tan.  J 

But r-x ^=7-: — 7^ ;  whence  ) 7^=)—. — ?:r6;  J  and =r 

tan.  B     tan.  .2-     (sm.  S)^'  (tan.  B)2     (sm.  S)2 '  tan.  B 

=  .  '     ,  or  sin.  S  :  sin.  D  : :  tan.  B  :  tan.  J,  that  is,  sin.  (C+B)  :  sin. 
sm.  o 

(C— B)  : :  tan.  J  BC  :  tan.  J  (AB— AC) ;  which  is  the  first  part  of  the 

...        tan.  ^     cos.  S  X  sin.  D        .  ,     tan.  2 

proposition.     Again,  since -=-^ — r; ^,  or  inversely -  = 

^    ^  ^      *  tan.  2     sm.  S  X  cos.  D  ^   tan.  J 

sin.  Sxcos.  D         ,    .        tan. X     sin.  Dxcos.  D      ,       -      ,         ,^.  ,. 

^ — -. — =- ;  and  since tt=-^ — f: ^  ;  therefore  by  multipli- 

cos.  S  X  sm.  D  tan.  B     sm.  D  X  cos.  S  '  J  r 

tan.  X     i2in.2     (cos.  D)2 


SPHERICAL  TRIGONOMETRY.  261 

„     .  •,•,■.■       1      tan.  X     tan.  S  x  tan.  zf 

But  It  was  already  shewn  that rr=^ — -. r^rrr — ,  wherefore  also 

•'  tan.  B  (tan.  B)^ 

tan.X     tan.  .2'_(tan.  sy 

tan.  B^tan.  ^"~(tan.  B)2* 

tan.  X     tan.  2     (cos.  Dp       ,       .       ,  , 

Now, =rX 7=7 ^,  as  has  just  been  shewn. 

tan.  B     tan.  J     (cos.  S)^ 

^,       ^      (cos  D)2     (tan.  2:f      .  ,    cos.  D     tan.  2 

Therefore  ) ^^=7; ^(2'^^'!  consequently ^-=-^ —,ot  cos. 

(cos.  S)2     (tan.  B)^  ^         -^  cos.  S     tan.  B 

S  :  cos.  D  : :  tan.  B  :  tan.  2,  that  is,  cos.  (C+B)  :  cos.  (C— B)  : :  tan.  ^ 

BC  :  tan.  J  (C+B) ;  which  is  the  second  part  of  the  proposition. 

Cor.  1.  By  applying  this  proposition  to  the  triangle  supplemental  to 
ABC  (11.)  and  by  considering,  that  the  sine  of  half  the  sum  or  half  the 
difference  of  the  supplements  of  two  arcs,  is  the  same  with  the  sine  of  half 
the  sum  or  half  the  difference  of  the  arcs  themselves  :  and  that  the  same 
is  true  of  the  cosines,  and  of  the  tangents  of  half  the  sum  or  half  the  dif- 
ference of  the  supplements  of  two  arcs  :  but  that  the  tangent  of  half  the 
supplement  of  an  arc  is  the  same  with  the  cotangent  of  half  the  arc  itself; 
it  will  follow,  that  the  sine  of  half  the  sum  of  any  two  sides  of  a  spherical 
triangle,  is  to  the  sine  of  half  their  difference  as  the  cotangent  of  half  the 
angle  contained  between  them,  to  the  tangent  of  half  the  difference  of  the 
angles  opposite  to  them  :  and  also  that  the  cosine  of  half  the  sum  of  these 
sides,  is  to  the  cosine  of  half  their  difference,  as  the  cotangent  of  half  the 
angle  contained  between  them,  to  the  tangent  of  half  the  sum  of  the  angles 
opposite  to  them. 

CoR.  2.  If  therefore  A,  B,  C,  be  the  three  angles  of  a  spherical  trian- 
gle, flf,  b,  c  the  sides  opposite  to  them, 

I.  sin.  ^  (A+B)  :  sin.  J  (A—B)  : ;  tan.^c  :  tan.i(a — b). 
II.  cos.  ^  (A+B)  :  COS.  J  (A—B)  : ;  tan.  |  c  :  tan.  I  (a-\-b). 

III.  sin.  I  (a+i)  :  sin.  J  (a— 5)  : :  tan.  ^  C  :  tan.  ^  (A—B). 

IV.  cos.  I  (a+b)  :  cos.  |  (a—b)  : :  tan.  |  C  :  tan.  -J  (A+B). 


262 


SPHERICAL  TRIGONOMETRY. 


PROBLEM  I. 

In  a  right  angled  spherical  triangle,  of  the  three  sides  and  three  angles,  any 
two  being  given,  besides  the  right  angle,  to  Jind  the  other  three. 

This  problem  has  sixteen  cases,  the  solutions  of  which  are  contained 
in  the  following  table,  where  ABC  is  any  spherical  triangle  right  angled 
at  A. 


GIVEN. 

SOUGHT. 

SOLUTION. 

BC  and  B. 

AC. 

AB. 

C. 

R  :  sin  BC  :  :  sin  B  :  sin  AC,   (19). 
R  :  cos  B  :  :  tan  BC  :  tan  AB,   (21). 
R  :  cos  BC  :  :  tan  B  :  cot  C,      (20). 

1 
2 
3 

AC  and  C. 

AB. 

BC. 

B. 

R  :  sin  AC  :  :  tan  C  :  tan  AB,  (18). 
cos  C  :  R  : :  tan  AC  :  tan  BC,  (21). 
R  :  cos  AC  : :  sin  C  :  cos  B,     (23). 

4 
5 
6 

AC  and  B. 

AB. 

BC. 

C. 

tan  B  :  tan  AC  : :  R  :  sin  AB,  (18). 
sin  B  :  sin  AC  : :  R  :  sin  BC,  (19). 
cos  AC  :  cos  B  :  :  R  :  sin  C,     (23). 

7 
8 
9 

AC  and  BC. 

AB. 
B. 
C. 

cos  AC  :  cos  BC  ; :  R  :  cos  AB,  (22). 
sin  BC  :  sin  AC  :  :  R  :  sin  B,     (19). 
tan  BC  :  tan  AC  : :  R  :  cos  C,    (21). 

10 
11 
12 

AB  and  AC. 

BC. 
B. 
C. 

R  :  cos  AB  ; :  cos  AC  :  cos  BC,  (22). 
sin  AB  :  R  : :  tan  AC  :  tan  B,     (18). 
sin  AC  :  R  : :  tan  AB  :  tan  C,     (18). 

13 

14 
14 

B  and  C. 

AB. 
AC. 
BC. 

sin  B  :  cos  C  : ;  R  :  cos  AB,     (23). 
sin  C  :  cos  B  : :  R  :  cos  AC,     (23). 
tan  B  :  cot  C  : :  R  :  cos  BC,      (20). 

15 
15 
16 

SPHERICAL  TRIGONOMETRY. 


2m 


TABLE  for  determining  the  affections  of  the  Sides  and  Angles  found  by 
the  preceding  rules. 


AC  and  B  of  the  same  affection. 

1 

If  BC^  90°,  AB  and  B  of  the  same  affection,  otherwise  dif- 

ferent,                                                                  (Cor.  15.) 

2 

If  BC/  90*^,  C  and  B  of  the  same  affection,  otherwise  diffe- 

rent,                                                                              (15.) 

3 

4 

AB  and  C  are  of  the  same  affection,                               (14.) 

If  AC  and  C  are  of  the  same  affection,  BC^^  90°  ;  otherwise 

BCZ90O,                                                           (Cor.  15.) 

5 

B  and  AC  are  of  the  same  affection,                               (14.) 

6 

Ambiguous. 

7 

Ambiguous. 

8 

Ambiguous. 

9 

When  BC/  90°,  AB  and  AC  of  the  same ;  otherwise  of  dif- 

ferent affection,                                                             (15.) 

10 

AC  and  B  of  the  same  affection,                                     (14.) 

11 

When  BC/90O,  AC  and  C  of  the  same  ;  otherwise  of  dif- 

ferent affection,                                                   (Cor.  15.) 

12 

BC^90o,  when  AB  and  AC  are  of  the  same  affection. 

(1.  Cor.  15.) 

13 

B  and  AC  of  the  same  affection,                                    (14.) 

14 

C  and  AB  of  the  same  affection,                                     (14.) 

14 

AB  and  C  of  the  same  affection,                                     (14.) 

15 

AC  and  B  of  the  same  affection,                                     (14.) 

15 

When  B  and  C  are  of  the  same  affection,  60/90°,  other- 

wise, BC  790°,                                                            (15.) 

16 

The  cases  marked  ambiguous  are  those  in  which  the  thing  sought  has 
two  values,  and  may  either  be  equal  to  a  certain  angle,  or  to  the  supple- 
ment of  that  angle.  Of  these  there  are  three,  in  all  of  which  the  things 
given  are  a  side,  and  the  angle  opposite  to  it ;  and  accordingly,  it  is  easy  to 
shew  that  two  right  angled  spherical  triangles  may  always  be  found  that 
have  a  side  and  the  angle  opposite  to  it  the  same  in  both,  but  of  which  the 
remaining  sides,  and  the  remaining  angle  of  the  one,  are  the  supplements 
of  the  remaining  sides  and  the  remaining  angle  of  the  other,  each  of  each. 

Though  the  affection  of  the  arc  or  angle  found  may  in  all  the  other  cases 
be  determined  by  the  rules  in  the  second  of  the  preceding  tables,  it  is  of 
use  to  remark,  that  all  these  rules  except  two,  may  be  reduced  to  one,  viz. 
that  when  the  thing  found  by  the  rules  in  the  first  table  is  either  a  tangent  or 
a  cosine ;  and  when,  of  the  tangents  or  cosines  employed  in  the  computation  of 
itf  one  only  belongs  to  an  obtuse  angle,  the  angle  required  is  also  obtuse. 


264  SPHERICAL  TRIGONOMETRY. 

Thus,  in  the  15th  case,  when  cos  AB  is  found,  if  C  be  an  obtuse  angle, 
because  of  cos  C,  AB  must  be  obtuse  ;  and  in  case  16,  if  either  B  or  C  b© 
obtuse,  BC  is  greater  than  90°,  but  if  B  and  C  are  either  both  acute,  or 
both  obtuse,  BC  is  less  than  90°. 

It  is  evident,  that  this  rule  does  not  apply  when  that  which  is  found  is 
the  sine  of  an  arc  ;  and  this,  besides  the  three  ambiguous  cases,  happens 
also  in  other  two,  viz.  the  1st  and  11th.  The  ambiguity  is  obviated,  in 
these  two  cases,  by  this  rule,  that  the  sides  of  a  spherical  right  angled  tri 
angle  are  of  the  same  affection  with  the  opposite  angles. 

Two  rules  are  therefore  sufficient  to  remove  the  ambiguity  in  all  the 
cases  of  the  right  angled  triangle,  in  which  it  can  possibly  bo  removed. 


SPHERICAL  TRIGONOMETRY. 


265 


It  may  be  useful  to  express  the  same  solutions  as  in  the  annexed  table. 
Let  A  be  at  the  right  angle  as  in  the  figure,  and  let  the  side  opposite  to  it 
be  a;  leib  be  th^  side  opposite  to  B,  and  c  the  side  opposite  to  C. 


GIVEN. 

SOUGHT. 

SOLUTION. 

1 
2 

3 
4 
5 
6 

a  and  B. 

b. 

c. 
C. 

sin  b  =  sm  a  X  sin  B. 
tanc  =tan  <z  X  cosB. 
cotC  =  cos  a  X  tan  B. 

b  and  C. 

c. 
a, 
B. 

tanc  =  sin  5  X  tan  C. 

tan  J 
tan  a  = 7-,* 

cosC 

cos  B  =  cos  6  X  sin  C. 

h  and  B. 

c, 
a. 
C. 

tan  b 
sm  c  =  - — =. 
tanB 

sin  6 
sm  a  =    .    ^. 
smB 

.    ^     cosB 
sinC= 1» 

cos  6 

7 
8 
9 

a  and  b. 

e, 
B. 
C. 

cos  a 

10 
11 
12 

cos  b 

.   „      sin  6 

sm  B  =  —. — . 
sm  a 

^      tan  b 
cos  C  = . 

tan  a 

b  and  c. 

a. 
B. 

C. 

cos  a  =  cos  b  X  cos  c. 

.      T>       tan  6 

tan  B  =  -; — . 

sm  c 

tan  c 

tan  C  =  -: r. 

sm  b 

13 
14 

14 

BandC. 

b 
a. 

cosC 

COSC  =-: -. 

sm  B 
cos  B 

cos  b  —    -r—^, 

smC 

cotC 

^^--tanB- 

15 
15 

34 


266 


SPHERICAL  TRIGONOMETRY. 


PROBLEM  II. 

In  any  oblique  angled  spherical  triangle^  of  the  three  sides  and  three  angles, 
any  three  being  given j  it  is  required  to  find  the  other  three. 

In  this  Table  the  references  (c.  4.),  (c.  5.),  &c.  are  to  the  cases  in  the 
preceding  Table,  (16.),  (27.),  &c.  to  the  propositions  in  Spherical  Trigo- 
nometry. 


GIVEN. 

SOUGHT. 

SOLUTION. 

1 

Two  sides 

AB,  AC, 

and  the  in- 

2  eluded  angle 

A. 

One  of  the 

other  angles 

B. 

Let  fall  the  perpendicular  CD  from 
the  unknown  angle,  not  requir- 
ed, on  AB. 

R  :  cos  A  :  ;  tan  AC  :  tan  AD, 
(c.  2.) ;  therefore  BD  is  known, 
and  sin  BD  :  sin  AD  : :  tan  A  : 
tan  B,  (27.) ;  B  and  A  are  of 
the  same  or  different  affection, 
according  as  AB  is  greater  or 
less  than  BD,  (16.). 

The  third 
side 
BC. 

Let  fall  the  perpendicular  CD  from 
one  of  the  unknown  angles  on 
the  side  AB. 

R  :  cos  A  :  :  tan  AC  :  tan  AD, 
(c.  2.) ;  therefore  BD  is  known, 
and  cos  AD  :  cos  BD  :  :  cos  AC 
:  cos  BC,  (26.) ;  according  as 
the  segments  AD  and  DB  are  of 
the  same  or  different  affection, 
AC  and  CB  will  be  of  the  same 
or  different  affection. 

SPERICAL  TRIGONOMETRY. 


267 


TABLE  continued. 


GIVEN. 

SOUGHT. 

SOLUTION. 

3 

Two  angles, 
A  and  ACB, 

and 

AC, 
the  side  be- 
tween them. 

4 

The  side 
BC. 

From  C  the  extremity  of  AC  near 
the  side  sought,  let  fall  the  per- 
pendicular CD  on  AB. 

R  :  cos  AC  :  :  tan  A  :  cot  ACD, 
(c.  3.) ;  therefore  BCD  is  known, 
and  cos  BCD  :  cos  ACD  :  :  tan 
AC  :  tan  BC,  (28.).  BC  is  less 
or  greater  than  90°,  according 
as  the  angles  A  and  BCD  are 
of  the  same,  or  different  atfec- 
tion. 

The  third 

angle 

B. 

Let  fall  the  perpendicular  CD  from 
one  of  the  given  angles  on  the 
opposite  side  AB. 

R  :  cos  AC  :  :  tan  A  :  cot  ACD, 
(c.  3.) ;  therefore  the  angle  BCD 
is  given,  and  sin  ACD  :  sin  BCD 
:  :  cos  A  :  cos  B,  (25.) ;  B  and 
A  are  of  the  same  or  differ- 
ent affection,  according  as  CD 
falls  within  or  without  the  tri- 
angle, that  is,  according  as  ACB 
is  greater  or  less  than  BCD, 
(16.). 

268 


SPHERICAL  TRIGONOMETRY. 


TABLE  continued. 


GIVEN. 

SOUGHT. 

SOLUTION. 

5 

Two  sides 

AC  and  BC, 

and  an  angle 

A 

opposite  to 
6 
one  of  them, 

BC. 

7 

0 

The  angle 

B 
opposite  to 
the  other  gi- 
ven side 
AC. 

Sin  BC  :  sin  AC  :  :  sin  A :  sin  B, 
(24.)     The  affection  of  B  is  am- 
biguous, unless  it  can  be  deter- 
mined by  this  rule,  that  accord- 
ing as  AC  -f  BC  is  greater  or 
less  than  180^,  A-fB  is  greater 
or  less  than  180°,  (10.) 

The  angle 

ACB 

contained  by 

the  given 

sides 

AC  and  BC. 

From  ACB  the  angle  sought  draw 
CD  perpendicular  to  AB  ;  then 
R  :  cos  AC  : :  tan  A  :  cot  ACD, 
(c.  3.) ;  and  tan  BC  :  tan  AC  ; : 
cos  ACD  ;  cos  BCD,  (28.)  ACD 
±  BCD  =  ACB,  and  ACB  is 
ambiguous,  because  of  the  am- 
biguous sign  -f  or  — . 

The  third 
side 
AB. 

Let  fall  the  perpendicular  CD  from 
the  angle  C,  contained  by  the 
given  sides,  upon  the  side  AB. 
R  :  cos  A  :  :  tan  AC  :  tan  AD, 
(c.  2.) ;  cos  AC  ;  cos  BC  :  :  cos 
AD  :  cos  BD,  (26.) 
AB=AD±BD,  wherefore  AB 
is  ambiguous. 

SPHERICAL  TRIGONOMETRY. 


269 


TABLE  continued. 


GIVEN. 

SOUGHT. 

SOLUTION. 

The  side 

Sin  B  :  sin  A  : :  sin  AC  :  sin  BC, 

BC 

(24) ;  the  affection  of  BC  is  un- 

opposite 

certain,  except  when  it  can  be  de- 

8 

to  the 

termined  by  this  rule,  that  accord- 

other 

ing  as  A+B  is  greater  or  less  than 

given  an- 

180<=>, AC+BC  is  also  greater  or 

gle  A. 

less  than  180°,  (10.). 

Two  angles 

From  the  unknown  angle  C,  draw 

A,B, 

The  side 

CD   perpendicular  to  AB  ;    then 

AB 

R  :  cos  A  : :  tan  AC  :  tan  AD, 

and  a  side 

adjacent 

(c.  2.) ;  tan  B  :  tan  A  :  :  sin  AD  : 

to  the 

sin  BD.     BD  is  ambiguous  ;  and 

9 

AC 

given 

therefore   AB  =  AD  i  BD  may 

angles 

have  four  values,  some  of  which 

opposite  to 

A,B. 

will  be  excluded  by  this  condition, 
that  AB  must  be  less  than  180°. 

one  of  them, 

From  the  angle  required,  C,  draw  CD 

B. 

perpendicular  to  AB. 

The  third 

R  :  cos  AC  : :  tan  A  :  cot  ACD, 
(c.  3.),  cos  A  :  cos  B  : :  sin  ACD  : 

angle 

sin  BCD,  (25.).     The  affection  of 

10 

BCD  is  uncertain,  and  therefore 

ACB. 

ACB  =  ACD  ±  BCD,  has  four 
values,  some  of  which  may  be  ex- 
cluded by  the  condition,  that  ACB 
is  less  than  180^. 

From  C  one  of  the  angles  not  requir- 

The three 

ed,  draw  CD  perpendicular  to  AB. 
Find  an  arc  E  such  that  tan  J  AB 

sides, 

:  tan  |  (AC-fBC) : :  tan  J  (AC- 
BC)  :  tan  J  E  ;  then,  if  AB  be 

11 

One  of  the 

AB,  AC, 

greater  than  E,  AB  is  the  sum,  and 

angles 

E  the  difference  of  AD  and  DB ; 

and 

but  if  AB  be  less  than  E,  E  is  the 

A. 

sum  and  AB  the  difference  of  AD, 

BC. 

DB,  (29.).  In  either  case,  AD  and 
BD  are  known,  and  tan  AC  :  tan 

AD  : :  R  :  cos  A. 

270 


SPHERICAL  TRIGONOMETRY. 


TABLE  continued. 


GIVEN. 

BOUGHT. 

SOLUTION. 

12 

The  three 

angles 

A,  B,  C. 

One  of  the 
sides 
BC. 

Suppose  the   supplements  of  the 
three  given  angles,  A,  B,  C,  to 
be  a,  bj  c,  and  to  be  the  sides  of 
a  spherical  triangle.     Find,  by 
the  last  case,  the  angle  of  this 
triangle,  opposite  to  the  side  a, 
and  it  will  be  the  supplement  of 
the  side  of  the  given  triangle  op- 
posite to  the  angle  A,  that  is,  of 
BC,  (11.) ;  and  therefore  BC  is 
found. 

In  the  foregoing  table,  the  rules  are  given  for  ascertaining  the  affection 
of  the  arc  or  angle  found,  whenever  it  can  be  done  :  Most  of  these  rules 
are  contained  in  this  one  rule,  which  is  of  general  application,  viz.  that 
when  the  thing  found  is  either  a  tangent  or  a  cosine^  and  of  the  tangents  or 
cosines  employed  in  the  computation  of  it,  either  one  or  three  belong  to  obtuse 
angles,  the  angle  found  is  also  obtuse.  This  rule  is  particdarly  to  be  attend- 
ed to  in  cases  5  and  7,  where  it  removes  part  of  the  ambiguity. 

It  may  be  necessary  to  remark  with  respect  to  the  11th  case,  that  the 
segments  of  the  base  computed  there  are  those  cut  off  by  the  nearest  per- 
pendicular; and  also,  that  when  the  sum  of  the  sides  is  less  than  180^, 
the  least  segment  is  adjacent  to  the  least  side  of  the  triangle ;  otherwise 
to  the  greatest,  (17.). 


SPHERICAL  TRIGONOMETRY. 


271 


The  last  table  may  also  be  conveniently  expressed  in  the  following 
manner,  denoting  the  side  opposite  to  the  angle  A,  by  a,  to  B  by  b,  and  to 
C  by  c ;  and  also  the  segments  of  the  base,  or  of  opposite  angle,  by  a; 
and  y. 


Two  sides 

b  and  c,  and 

the  angle 

between 

them  A. 


Angles 

A  and  C 

and 

side  b 


Sides 
a  and  6 

and 
angle  A. 


B 


B 


Find  Xj  so  that 
tan  a;=tan  b  X  cos  A ;  then 
^      _     sin  a;  X  tan  A 

tan  B= — : — ; r-- 

sm  (c— a;) 


Find  Xf  as  above, 

o  cos  b  X  cos  (c^x) 

then  cos  a=z i .', 

cos  X 


Find  X,  so  that 
cot  a; = cos  J  X  tan  A ;  then 
tan  b  X  cos  x 


tan  a: 


cos  (c—x) 


Find  Xf  as  above, 

.             _     cos  A  X  sin  (c^x) 
then  cos  B= : ^ ■'. 


sma; 


sin  B= 


sin  ^X  sin  A 


sma 


Find  a?,  so  that 

cot  ir=cos  ixtan  A  ;  then 

^     cosa;Xtan  b 

cos  C= . 

tan  a 


Find  Xf  so  that 

tana:=tan5xcos  A;  and  find 

y,  so  that 

cos  cXcos  X 
cos  y=s 


cos  b 


■=X:k!/' 


272 


SPHERICAL  TRIGONOMETRY. 


TABLE  continued. 


SOLOTION. 


10 


The  angles 

AandB 

and  the 

side  b. 


sma=: 


sin  J  X  sin  A 


sin  B 


Find  Xf  so  that 

tan  a:=tan  b  X  cos  A ;  and  y,  so 

that 

sin  itXtan  A 
smy= 


tan  B 


c^zxJi^i/. 


Find  a?,  so  that 

cot  a?=cos  b  X  tan  A ;  and  also  y, 

80  that 

sin  jtXcos  B 
sm  y== 


cos  A 


CrssffJ-y. 


11 


a,  b,  e. 


Let  a4-^+c=J. 


sm 


1 .  _  Vsin  (^^—Z^jXsin  (^s-^c) 
■v/sin  ^  X  sin  c 


or  cos  }A 


_  -y/sin  ^s  X  sin  (js—a) 


^sin  ixsin  c 


12 


A,  B,  C« 


Let  A-|-B+C=S. 


.    ,        Vcos  i  S  X  cos  (h  S— A) 
Vsin  B  X  sin  C 


orcosja 


_  v^cos(^S— B)lcos(S-C) 


Vsin  B  X  sin  C 


APPENDIX 


TO 


SPHERICAL 

TRIGONOMETRY, 


CONTAINING 


NAPIER'S  RULES  OF  THE  CIRCULAR  PARTS. 


The  rule  of  the  Circular  Parts,  invented  by  Napier,  is  of  great  use  in 
Spherical  Trigonometry,  by  reducing  all  the  theorems  employed  in  the 
solution  of  right  angled  triangles  to  two.  These  two  are  not  new  proposi- 
tions, but  are  merely  enunciations,  which,  by  help  of  a  particidar  arrange- 
ment and  classification  of  the  parts  of  a  triangle,  include  all  the  six  propo- 
sitions, with  their  corollaries,  which  have  been  demonstrated  above  from 
the  i8th  to  the  23d  inclusive.  They  are  perhaps  the  happiest  example  of 
artificial  memory  that  is  known. 


DEFINITIONS. 


1.  If  in  a  spherical  triangle,  we  set  aside  the  right  angle,  and  consider  only 
the  five  remaining  parts  of  the  triangle,  viz.  the  three  sides  and  the  two 
oblique  angles,  then  the  two  sides  which  contain  the  right  angle,  and 
the  complements  of  the  other  three,  namely,  of  the  two  angles  and  the 
hypotenuse,  are  called  the  Circular  Parts. 

Thus,  in  the  triangle  ABC  right  angled  at  A,  the  circular  parts  are  AC, 
AB  with  the  complements  of  B,  BC,  and  C.     These  parts  are  called 
circular ;  because,  when  they  are  named  in  the  natural  order  of  tl 
succession,  they  go  round  the  triangle. 

2.  When  of  the  five  circular  parts  any  one  is  taken,  for  the  middle  par 
then  of  the  remaining  four,  the  two  which  are  immediately  adjacent  to 
it,  on  the  right  and  left,  are  called  the  adjacent  parts  ;  and  the  other  two, 
each  of  which  is  separated  from  the  middle  by  an  adjacent  part,  are  call- 
ed opposite  parts. 

Thus  in  the  right  angled  triangle  ABC,  A,  being  the  right  angle,  AC,  AB, 
90O— B,  90O— BC,  90O— C,  are  the  circular  parts,  by  Def.  1. ;  and  if 

Z5 


274  APPENDIX  TO 

any  one,  as  AC,  be  reckoned  the  middle  part,  then  AB  and  90<^— C,  which 
are  contiguous  to  it  on  different  sides,  are  called  adjacent  parts  ;  and  90° 
— B,  90°— BC  are  the  opposite  parts.     In  like  manner  if  AB  is  taken  for 


the  middle  part,  AC  and  90°— B  are  the  adjacent  parts  :  90^— BC,  and 
90<^— C  are  the  opposite.  Or  if  90O--BC  be  the  middle  part,  90--B, 
90° — C  are  adjacent  ;  AC  and  AB  opposite,  <&c. 

This  arrangement  being  made,  the  rule  of  the  circular  part  is  contained 
in  the  following 

PROPOSITION. 

In  a  right  angled  spherical  triangle,  the  rectangle  under  the  radius  and  the  sine 
of  the  middle  part,  is  equal  to  the  rectangle  under  the  tangents  of  the  adjacent 
parts ;  or,  to  the  rectangle  under  the  cosines  of  the  opposite  parts 

The  truth  of  the  two  theorems  included  in  this  enunciation  may  be 
easily  proved,  by  taking  each  of  the  five  circular  parts  in  succession  for 
the  middle  part,  when  the  general  proposition  will  be  found  to  coincide 
with  some  one  of  the  analogies  in  the  table  already  given  for  the  resolution 
of  the  cases  of  right  angled  spherical  triangles.  Thus,  in  the  triangle  ABC, 
if  the  complement  of  the  hypotenuse  BC  be  taken  as  the  middle  part,  90° 
— iB,  and  90° — C,  are  the  adjacent  parts,  AB  and  AC  the  opposite.  Then 
the  general  rule  gives  these  two  theorems,  Rxcos  BC— cot  Bxcot  C, 
'and  Rx  cos  BC=cos  AB  X  cos  AC.  The  former  of  these  coincides  with 
the  cor.  to  the  20th  ;  and  the  latter  with  the  22d. 

To  apply  the  foregoing  general  proposition  to  resolve  any  case  of  a  right 
angled  spherical  triangle,  consider  which  of  the  three  qualities  named 
(the  two  things  given  and  the  one  required)  must  be  made  the  middle  term, 
in  order  that  the  other  two  may  be  equi-distant  from  it,  that  is,  may  be 
both  adjacent,  or  both  opposite  ;  then  one  or  other  of  the  two  theorems 
contained  in  the  above  enunciation  will  give  the  value  of  the  thing  re- 
quired. 

Suppose,  for  example,  that  AB  and  BC  are  given,  to  find  C ;  it  is  evi- 
dent that  if  AB  be  made  the  middle  part,  BC  and  C  are  the  opposite  parts, 
and  therefore  Rxsin  AB=sin  Cxsin  BC,  for  sin  C=cos  (90°— C),  and 

cos  (90°— BC)=sin  BC,  and  consequently  sin  C=- — ^. 

Again,  suppose  that  BC  and  C  are  given  to  find  AC ;  it  is  obvious  that 
C  is  in  the  middle  between  the  adjacent  parts  AC  and  (90° —BC),  there- 


SPHERICAL  TRIGONOMETRY.  275 

fore  R  X  cos  C=tan  AC  X  cot  BC,  or  tan  AC= ^7^=cos  C  +  tan  BC  •, 

cot  i>L/ 

because,  as  has  been  shewn  above, =;^=:tan  BC. 

cot  BC 

In  the  same  way  may  all  the  other  cases  be  resolved.  One  or  two  trials 
will  always  lead  to  the  knowledge  of  the  part  which  in  any  given  case  is 
to  be  assumed  as  the  middle  part ;  and  a  little  practice  will  make  it  easy, 
even  without  such  trials,  to  judge  at  once  which  of  them  is  to  be  so  as- 
sumed. It  may  be  useful  for  the  learner  to  range  the  names  of  the  five 
circular  parts  of  the  triangle  round  the  circumference  of  a  circle,  at  equal 
distances  from  one  another,  by  which  means  the  middle  part  will  be  imme- 
diately determined. 

Besides  the  rule  of  the  circular  parts,  Napier  derived  from  the  last  of  the 
three  theorems  ascribed  to  him  above,  (schol.  29.)  the  solutions  of  all  the 
cases  of  oblique  angled  triangles.  These  solutions  are  as  follows  :  A,  B, 
C,  denoting  the  three  triangles  of  a  spherical  triangle,  and  a,  5,  c,  the  sides 
opposite  to  them. 

I. 

Given  two  sides  i,  c,  and  the  angle  A  between  them. 
To  find  the  angles  B  and  C. 

uni(B-C)=cotiAx!£iigl.    (31.)cor.l. 
.aa|(B+C)=cotJAx^2i||^.    (31.)  cor.  1. 

To  find  the  third  side  a. 
sin  B  :  sin  A  : :  sin  h  :  sin  a. 

XL 

Given  the  two  sides  5,  c,  and  the  angle  B  opposite  to  one  of  them. 

To  find  C,  and  the  angle  opposite  to  the  other  side. 

sin  6  :  sin  c  : :  sin  B  :  sin  C. 

To  find  the  contained  angle  A. 
co.iA=tani(B-C)xSl|ig|.    (31.)  cor.  1. 

To  find  the  third  side  a. 
sin  B  :  sin  A  : :  sin  h  :  sin  a. 

III. 

Given  two  angles  A  and  B,  and  the  side  c  between  them. 
To  find  the  other  two  sides  a,  h. 


276  APPENDIX  TO 

Uni(»-<.)=ta„JoX^^iiArB}.     (31.) 

.an^(i+«)=ta„i.x:-2iii|^.    (31.) 

To  find  the  third  angle  C. 
sin  a :  sin  c  : :  sin  A  :  sin  C. 


IV. 

Given  two  angles  A  and  B,  and  the  side  cr,  opposite  to  one  of  them. 

To  find  h,  the  side  opposite  to  the  other. 

sin  A  :  sin  B  :  :  sin  a :  sin  b. 

To  find  c,  the  side  between  the  given  angles. 

«.ic=.anH<.-i)xS^|^.    (31.) 

To  find  the  third  angle  C. 
sin  a :  sin  c  : :  sin  A  :  sin  C. 

The  other  two  cases,  when  the  three  sides  are  given  to  find  the  angles, 
or  when  the  three  angles  are  given  to  find  the  sides,  are  resolved  by  the 
29th,  (the  first  of  Napier's  Propositions,)  in  the  same  way  as  in  the  table 
already  given  for  the  case  of  the  oblique  angled  triangle. 

There  is  a  solution  of  the  case  of  the  three  sides  being  given,  which  it 
is  often  very  convenient  to  use,  and  which  is  set  down  here,  though  the 
proposition  on  which  it  depends  has  not  been  demonstrated. 

Let  Oy  b,  c,  be  the  three  given  sides,  to  find  the  angle  A,  contained  be- 
tween b  and  c. 


If  Rad  =  1,  and  o  +  5  +  c  =  ^, 

sin  " 


j^  -^  ysin  (^  s-b)  X  sin  ^  (s^c)  ^  ^^^ 
•/sin  6  X  sin  c 


cos  i  A  ^V«i"-(^^Xsin^(.-a)) 
^sin  5  X  sin  c 

In  like  manner,  if  the  three  angles,  A,  B,  C  are  given  to  find  c,  the  side 
between  A  and  B. 


SPHERICAL  TRIGONOMETRY.  277 

S. 


LetA  +  B+C  = 

.    J         ycos  ^  S  X  cos  (^  S  —A) 

sm  "2  c —  ■:= f  or, 

V^sin  B  X  sin  C 

cos  1  c=  ^^^'  (^  S^BHTcos  (1  S~C) 
^  ^sin  B  X  sin  C. 


These  theorems,  on  account  of  the  facility  with  which  Logarithms  are 
applied  to  them,  are  the  most  convenient  of  any  for  resolving  the  two  cases 
to  which  they  refer.  When  A  is  a  very  obtuse  angle,  the  second  theorem, 
which  gives  the  value  of  the  cosine  of  its  half,  is  to  be  used ;  otherwise 
the  first  theorem,  giving  the  value  of  the  sine  of  its  half  its  preferable. 
The  same  is  to  be  observed  with  respect  to  the  side  c,  the  reason  of  which 
was  explained.  Plane  Trig.  Schol. 


END  OF  SPHERICAL  TRIGONOMETRY. 


1^ 


NOTES 

ON    THE 

FIRST  BOOK  OF  THE  ELEMENTS. 


DEFINITIONS. 
I. 


In  the  definitions  a  few  changes  have  been  made,  of  which  it  is  neces- 
sary to  give  some  account.  One  of  these  changes  respects  the  first  defini- 
tion, that  of  a  point,  which  Euclid  has  said  to  be,  *  That  which  has  no 
parts,  or  which  has  no  magnitude.'  Now,  it  has  been  objected  to  this  defi- 
nition, that  it  contains  only  a  negative,  and  that  it  is  not  convertible,  as 
every  good  definition  ought  certainly  to  be.  That  it  is  not  convertible  is 
evident,  for  though  every  point  is  unextended,  or  without  magnitude,  yet 
every  thing  unextended  or  without  magnitude,  is  not  a  point.  To  this  it 
is  impossible  to  reply,  and  therefore  it  becomes  necessary  to  change  the 
definition  altogether,  which  is  accordingly  done  here,  a  point  being  defined 
to  be,  that  which  has  position  hut  not  magnitude.  Here  the  affirmative  part 
includes  all  that  is  essential  to  a  point,  and  the  negative  part  includes 
every  thing  that  is  not  essential  to  it.  I  ani  indebted  for  this  definition  to 
a  friend,  by  whose  judicious  and  learned  remarks  I  have  often  profited. 


II. 

After  the  second  definition  Euclid  has  introduced  the  following,  "  the 
"  extremities  of  aline  are  points." 

Now,  this  is  certainly  not  a  definition,  but  an  inference  from  the  defini- 
tions of  a  point  and  of  aline.  That  which  terminates  aline  can  have  no 
breadth,  as  the  line  in  which  it  is  has  none  ;  and  it  can  have  no  length,  as  it 
would  not  then  be  a  termination,  but  a  part  of  that  which  is  supposed  to 
terminate.  The  termination  of  a  line  can  therefore  have  no  magnitude,  and 
having  necessarily  position,  it  is  a  point.  But  as  it  is  plain,  that  in  all  this 
we  are  drawing  a  consequence  from  two  definitions  already  laid  down,  and 
not  giving  a  new  definition,  I  have  taken  the  liberty  of  putting  it  down  as 
a  corollary  to  the  second  definition,  and  have  added,  that  the  intersections  of 
one  line  with  another  are  points,  as  this  affords  a  good  illustration  of  the  nature 
of  a  point,  and  is  an  inference  exactly  of  the  same  kind  with  the  preceding. 
The  same  thing  nearly  has  been  done  with  the  fourth  definition,  where 
that  which  Euclid  gave  as  a  separate  definition  is  made  a  corollary  to  the 


280 


NOTES. 


fourth,  because  it  is  in  fact  an  inference  deduced  from  comparing  the  defi- 
nitions of  a  superficies  and  a  line. 

As  it  is  impossible  to  explain  the  relation  of  a  superficies,  a  line,  and  a 
point  to  one  another,  and  to  the  solid  in  which  they  all  originate,  better 
than  Dr.  Simson  has  done,  I  shall  here  add,  with  very  little  change,  the 
illustration  given  by  that  excellent  Geometer. 

"  It  is  necessary  to  consider  a  solid,  that  is,  a  magnitude  which  has 
length,  breadth,  and  thickness,  in  order  to  understand  aright  the  definitions 
of  a  point,  line  and  superficies  ;  for  these  all  arise  from  a  solid,  and  exist  in 
it ;  The  boundary,  or  boundaries  which  contain  a  solid,  are  called  superfi- 
cies, or  the  boundary  which  is  common  to  two  solids  which  are  contiguous, 
or  which  divides  one  solid  into  two  contiguous  parts,  is  called  a  superfi- 
cies ;  Thus,  if  BCGF  be  one  of  the  boundaries  which  contain  the  solid 
ABCDEFGH,  or  which  is  the  common  boundary  of  this  solid,  and  the  solid 
BKLCFNMG,  and  is  therefore  in  the  one  as  well  as  the  other  solid,  it  is 
called  a  superficies,  and  has  no  thickness  ;  For  if  it  have  any,  this  thick- 
ness must  either  be  a  part  of  the  thickness  of  the  solid  AG,  or  the  soHd  BM, 
or  a  part  of  the  thickness  of  each  of  them.  It  cannot  be  a  part  of  the  thick- 
ness of  the  solid  BM  ;  because,  if  this  solid  be  removed  from  the  solid  AG, 
the  superficies  BCGF,  the  boundary  of  the  solid  AG,  remains  still  the 
same  as  it  was.  Nor  can  it  be  a  part  of  the  thickness  of  the  solid  AG : 
because  if  this  be  removed  from  the  solid  BM,  the  superficies  BCGF,  the 
boundary  of  the  solid  BM,  does  nevertheless  remain ;  therefore  the  super- 
ficies BCGF  has  no  thickness,  but  only  length  and  breadth. 

"  The  boundary  of  a  superficies  is  called  a  line  j  or  a  line  is  the  common 
boimdary  of  two  superficies  that  are  contiguous,  or  it  is  that  which  divides 
one  superficies  into  two  contiguous  parts  :  Thus,  if  BC  be  one  of  the  boun- 
daries which  contain  the  superficies  ABCD,  or  which  is  the  common  boun- 
dary of  this  superficies,  and  of  the  superficies  KBCL,  which  is  contiguous 
to  it,  this  boundary  BC  is  called  a  line,  and  has  no  breadth ;  For,  if  it  have 
any,  this  must  be  part  either  of  the  breadth  of  the  superficies  ABCD  or 
of  the  superficies  KBCL,  or  part  of 


each  of  them.  It  is  not  part  of  the 
breadth  of  the  superficies  KBCL ; 
for  if  this  superficies  be  removed  from 
the  superficies  ABCD,  the  line  BC 
which  is  the  boundary  of  the  super- 
ficies ABCD  remains  the  same  as  it 
was.  Nor  can  the  breadth  that  BC 
is  supposed  to  have,  be  a  part  of  the 
breadth  of  the  superficies  ABCD ;  be- 
cause, if  this  be  removed  from  the  su- 
perficies KBCL,  the  line  BC,  which 
is  the  boundary  of  the   superficies 


a 


p 


^c 


B 


KBCL,  does  nevertheless  remain :  Therefore  the  line  BC  has  no  breadth. 
And  because  the  line  BC  is  in  a  superficies,  and  that  a  superficies  has  no 
thickness,  as  was  shown ;  therefore  a  line  has  neither  breadth  nor  thick- 
ness, but  only  length. 

*'  The  boundary  of  a  line  is  called  a  point,  or  a  point  is  a  common  boun- 
dary or  extremity  of  two  lines  that  are  contiguous  :  Thus,  if  B  be  the  ex- 


NOTES 


281 


tremity  of  the  line  AB,  or  tlie  common  extremity  of  the  two  lines  AB,KB, 

this  extremity  is  called  a  point,  and  has  no  length :  For  if  it  have  any,  this 

length  must  either  be  part  of  the 

length  of  the  line  AB,  or  of  the  line 

KB.     It  is  not  part  of  the  length  of 

KB  ;  for  if  the  line  KB  be  removed 

from  AB,  the  point  B,  which  is  the 

extremity  of  the  line  AB,  remains  the 

same  as  it  was  ;  Nor  is  it  part  of  the 

length  of  the  line  AB  ;  for  if  A  B  be 

removed  from  the  line  KB,  the  point 

B,  which  is  the  extremity  of  the  line 

KB,    does    nevertheless    remain  : 

Therefore  the  point  B  has  no  length ; 

And  because  a  point  is  in  a  line,  and 

a  line  has  neither  breadth  nor  thickness,  therefore  a  point  has  no  length, 

breadth,  nor  thickness.     And  in  this  manner  the  definition  of  a  point,  line, 

and  superficies  are  to  be  understood." 


II 

G 

]W 

E 

A 

A 

A 

; 

P 

/ 

/ 

A 

B 

. 

[C 

III. 

Euclid  has  defined  a  straight  line  to  be  a  line  which  (as  we  translate  it) 
"lies  evenly  between  its  extreme  points."  This  definition  is  obviously 
faulty,  the  word  evenly  standing  as  much  in  need  of  an  explanation  sfs  the 
word  straight,  which  it  is  intended  to  define.  In  the  original,  however,  it 
must  be  confessed,  that  this  inaccuracy  is  at  least  less  striking  than  in  our 
translation ;  for  the  word  which  we  render  evenly  is  e^Lao,  equally^  and  is  ac- 
cordingly translated  ex  (squo,  and  equaliter  by  Comraandine  and  Gregory. 
The  definition,  therefore,  is,  that  a  straight  line  is  one  which  lies  equally 
between  its  extreme  points  :  and  if  by  this  we  understand  a  line  that  lies 
between  its  extreme  points  so  as  to  be  related  exactly  alike  to  the  space 
on  the  one  side  of  it,  and  to  the  space  on  the  other,  we  have  a  definition 
that  is  perhaps  a  little  too  metaphysical,  but  which  certainly  contains  in  it 
the  essential  character  of  a  straight  line.  That  Euclid  took  the  definition 
in  this  sense,  however,  ,is  not  certain,  because  he  has  not  attempted  to 
deduce  from  it  any  property  whatsoever  of  a  straight  line  ;  and  indeed,  it 
should  seem  not  easy  to  do  so,  without  employing  some  reasonings  of  a 
more  metaphysical  kind  than  he  has  any  where  admitted  into  his  Elements. 
To  supply  the  defects  of  his  definition,  he  has  therefore  introduced  the 
Axiom,  that  tvoo  straight  lines  cannot  inclose  a  space ;  on  which  Axiom  it  is, 
and  not  on  his  definition  of  a  straight  line,  that  his  demonstrations  are 
founded.  As  this  manner  of  proceeding  is  certainly  not  so  regular  and 
scientific  as  that  of  laying  down  a  definition,  from  which  the  properties  of 
the  thing  defined  maybe  logically  deduced,  I  have  substituted  another  defi- 
nition of  a  straight  line  in  the  room  of  Euclid's.  This  definition  of  a  straight 
line  was  suggested  by  a  remark  of-  Boscovich,  who,  in  his  Notes  on  the 
philosophical  Poem  of  Professor  Stay,  says,  "  Rectam  lineam  rectae  con- 
"  gruere  totam  toti  in  infinitum  productum  si  bina  puncta  unius  binis  al- 
"  terius  congruant,  patet  ex  ipsa  admodum  clara  rectitudinis  idea  quam 

36 


J82  NOTES. 

"habemus."  (Supplementum  m  lib.  3.  §  550.)  Now,  tbat  which  Mr. 
Boscovich  would  consider  a.s  an  inference  from  our  idea  of  straightness, 
seems  itself  to  be  the  essence  of  that  idea,  and  to  afford  the  best  criterion 
for  judging  whether  any  given  line  be  straight  or  not.  On  this  principle 
we  have  given  the  definition  above,  If  there  he  two  lines  which  cannot  coin- 
cide in  two  points  J  without  coinciding  altogether  ^  each  of  them  is  called  a  straight 
line.  I 

This  definition  was  otherwise  expressed  in  the  two  former  editions :  it 
was  said,  that  lines  are  straight  lines  which  cannot  coincide  in  part,  with- 
out coinciding  altogether.  This  was  liable  to  an  objection,  viz.  that  it  de- 
fined straight  lines,  but  not  a  straight  line ;  and  though  this  in  truth  is  but 
a  mere  cavil,  it  is  better  to  leave  no  room  for  it.  The  definition  in  the  form 
.  now  given  is  also  more  simple. 

From  the  same  definition,  the  proposition  which  Euclid  gives  as  an 
Axiom,  that  two  straight  lines  cannot  inclose  a  space,  follows  as  a  neces- 
sary consequence.  For,  if  two  lines  inclose  a  space,  they  must  intersect 
one  another  in  two  points,  and  yet,  in  the  intermediate  part,  must  not  coin- 
cide ;  and  therefore  by  the  definition  they  are  not  straight  lines.  It  follows 
in  the  same  way,  that  two  straight  lines  cannot  have  a  common  segment, 
or  cannot  coincide  in  part,  without  coinciding  altogether. 

After  laying  down  the  definition  of  a  straight  line,  as  in  the  first  Edition, 
I  was  favoured  by  Dr.  Reid  of  Glasgow  with  the  perusal  of  a  MS.  contain- 
ing many  excellent  observations  on  the  first  Book  of  Euclid,  such  as  might 
be  expected  from  a  philosopher  distinguished  for  the  accuracy  as  well  as 
the  extent  of  his  knowledge.  He  there  defined  a  straight  line  nearly  as 
has  been  done  here,  viz.  "  A  straight  line  is  that  which  cannot  meet  ano- 
"  ther  straight  line  in  more  points  than  one,  otherwise  they  perfectly  coincide, 
"  and  are  one  and  the  same."  Dr.  Reid  also  contends,  that  this  must  have 
been  Euclid's  own  definition  ;  because,  in  the  first  proposition  of  the 
eleventh  Book,  that  author  argues,  "  that  two  straight  lines  cannot  have  a 
"  common  segment,  for  this  reason,  that  a  straight  line  does  not  meet  a 
"  straight  line  in  more  points  than  one,  otherwise  they  coincide."  Whether 
this  amounts  to  a  proof  of  the  defilnition  above  having  been  actually 
Euclid's,  I  will  not  take  upon  me  to  decide ;  but  it  is  certainly  a  proof 
that  the  writings  of  that  Geometer  ought  long  since  to  have  suggested  this 
definition  to  his  commentators  ;  and  it  reminds  me,  that  I  might  have  learn- 
ed from  these  writings  what  I  have  acknowledged  above  to  be  derived  from 
a  remoter  source. 

There  is  another  characteristic,  ^d  obvious  property  of  straight  lines, 
by  which  1  have  often  thought  that  they  might  be  very  conveniently  defin- 
ed, viz.  that  the  position  of  the  whole  of  a  straight  line  is  determined  by  the 
position  of  two  of  its  points,  in  so  much  that,  when  two  points  of  a  straight 
line  continue  fixed,  the  line  itself  cannot  change  its  position.  It  might 
therefore  be  said,  that  a  straight  line  is  one  in  which,  if  the  position  of  two 
points  be  determined,  the  position  of  the  whole  line  is  determined.  But  this  de- 
finition, though  it  amount  in  fact  to  the  same  thing  with  that  already  given, 
is  rather  more  abstract,  and  not  so  easily  made  the  foundation  of  reason- 
ing. I  therefore  thought  it  best  to  lay  it  aside,  and  to  adopt  the  definition 
given  in  the  text. 


NOTES.  2^ 


V. 


The  definition  of  a  plane  is  given  from  Dr.  Simson,  Euclid's  being  liable 
to  the  same  objections  with  his  definition  of  a  straight  line  ;  for,  he  says, 
that  a  plane  superficies  is  one  which  "  lies  evenly  between  its  extreme 
"  Knes."  The  defects  of  this  definition  are  completely  removed  in  that  which 
Dr.  Simson  has  given.  Another  definition  different  from  both  might  have 
been  adopted,  viz.  That  those  superficies  are  called  plane,  which  are  such, 
that  if  three  points  of  the  one  coincide  with  three  points  of  the  other,  the 
whole  of  the  one  must  coincide  with  the  whole  of  the  other.  This  defini- 
tion, as  it  resembles  that  of  a  straight  line,  already  given,  might,  perhaps, 
have  been  introduced  with  some  advantage  ;  but  as  the  purposes  of  demon- 
stration cannot  be  better  answered  than  by  that  in  the  text,  it  has  been 
thought  best  to  make  no  farther  alteration. 


YI. 

In  Euclid,  the  general  definition  of  a  plane  angle  is  placed  before  that  of 
a  rectilineal  angle,  and  is  meant  to  comprehend  those  angles  which  are 
formed  by  the  meeting  of  the  other  lines  than  straight  lines.  A  plane 
angle  is  said  to  be  "the  inclination  of  two  lines  to  one  another  which 
"  meet  together,  but  are  not  in  the  same  direction."  This  definition  is 
omitted  here,  because  that  the  angles  formed  by  the  meeting  of  curve  lines, 
though  they  may  become  the  subject  of  geometrical  investigation,  certainly 
do  not  belong  to  the  Elements  ;  for  the  angles  that  must  first  be  considered 
are  those  made  by  the  intersection  of  straight  lines  with  one  another. 
The  angles  formed  by  the  contact  or  intersection  of  a  straight  line  and  a 
circle,  or  of  two  circles,  or  two  curves  of  any  kind  with  one  another, 
could  produce  nothing  but  perplexity  to  beginners,  and  cannot  possibly  be 
understood  till  the  properties  of  rectilineal  angles  have  been  fully  explained. 
On  this  ground,  I  am  of  opinion,  that  in  an  elementary  treatise  it  may 
fairly  be  omitted  Whatever  is  not  useful,  should,  in  explaining  the  ele- 
ments of  a  science,  be  kept  out  of  sight  altogether ;  for,  if  it  does  not  assist 
the  progress  of  the  understanding,  it  will  certainly  retard  it 


AXIOMS. 

Among  the  Axioms  there  have  been  made  only  two  alterations.  The 
10th  Axiom  in  Euclid  is,  that  "  two  straight  lines  cannot  inclose  a  space  ;" 
which,  having  become  a  corollary  to  our  definition  of  a  straight  line,  ceases 
of  course  to  be  ranked  with  self-evident  propositions.  It  is  therefore  re- 
moved from  among  the  Axioms. 

The  12th  Axiom  of  Euclid  is,  that  "  if  a  straight  line  meets  two  straight 
"  lines,  so  as  to  make  the  two  interior  angles  on  the  same  side  of  it  taken 
"  together  less  than  two  right  angles,  these  straight  lines  being  continually 
"  produced,  shall  at  length  meet  upon  that  side  on  which  are  the  angles 


284  NOTES. 

"which  are  less  than  two  right  angles."  Instead  of  this  proposition, 
which,  though  true,  is  by  no  means  self-evident ;  another  that  appeared 
more  obvious,  and  better  entitled  to  be  accounted  an  Axiom,  has  been  in- 
troduced, viz.  "  that  two  straight  lines,  which  intersect  one  another,  can- 
"not  be  both  parallel  to  the  same  straight  line."  On  this  subject,  how- 
ever, a  fuller  explanation  is  necessary,  for  which  see  the  note  on  the  29th 
Prop. 

PROP.  IV.  and  VIII.  B.  I. 

The  IV.  and  VIII.  propositions  of  the  first  book  are  tho  foimdation  of  all 
that  follows  with  respect  to  the  comparison  of  triangles.  They  are  de- 
monstrated by  what  is  called  the  method  of  superaposition,  that  is,  by  lay- 
ing the  one  triangle  upon  the  other,  and  proving  that  they  must  coincide. 
To  this  some  objections  have  been  made,  as  if  it  were  ungeometrical  to 
suppose  one  figure  to  be  removed  from  its  place  and  applied  to  another 
figure.  "  The  laying,"  says  Mr.  Thomas  Simson  in  his  Elements,  "  of 
"  one  figure  upon  another,  whatever  evidence  it  may  afford,  is  a  mechanical 
"  consideration,  and  depends  on  no  postulate."  It  is  not  clear  what  Mr. 
Simson  meant  here  by  the  word  mechanical :  but  he  probably  intended  only 
to  say,  that  the  method  of  superaposition  involves  the  idea  of  motion,  which 
belongs  rather  to  mechanics  than  geometry ;  for  I  think  it  is  impossible 
that  such  a  Geometer  as  he  was  could  meaa  to  assert,  that  the  evidence 
derived  from  this  method  is  like  that  which  arises  from  the  use  of  instru- 
ments, and  of  the  same  kind  with  what  is  furnished  by  experience  and  ob- 
servation. The  demonstrations  of  the  fourth  and  eighth,  as  they  are  given 
by  Euclid,  are  as  certainly  a  process  of  pure  reasoning,  depending  solely 
on  the  idea  of  equality,  as  established  in  the  8th  Axiom,  as  any  thing  in 
geometry.  But,  if  still  the  removal  of  the  triangle  from  its  place  be  consi- 
dered as  creating  a  difiiculty,  and  as  inelegant,  because  it  involves  an  idea, 
that  of  motion,  not  essential  to  geometry,  this  defect  may  be  entirely  re- 
medied, provided  that,  to  Euclid's  three  postulates,  we  be  allowed  to  add 
the  following,  viz.  That  if  there  he  two  equal  straight  lines,  and  if  any  figure 
whatsoever  he  constituted  on  the  one,  a  figure  every  way  equal  to  it  may  he  con- 
stituted on  the  other.  Thus  if  AB  and  DE  be  two  equal  straight  lines,  and 
ABC  a  triangle  on  the  base  AB,  a  triangle  DEF  every  way  equal  to  ABC 
may  be  supposed  to  be  constituted  on  DE  as  a  base.  By  this  it  is  not 
meant  to  assert  that  the  method  of  describing  the  triangle  DEF  is  actually 
known,  but  merely  that  the  triangle  DEF  may  be  conceived  to  exist  in 
all  respects  equal  to  the  triangle  ABC.  Now,  there  is  no  truth  whatso- 
ever that  is  better  entitled  than  this  to  be  ranked  among  the  Postulates  or 
Axioms  of  geometry  ;  for  the  straight  lines  AB  and  DE  being  every  way 
equal,  there  can  be  nothing  belonging  to  the  one  that  may  not  also  belong 
to  the.  other. 

On  the  strength  of  this  Postulate  the  IV.  proposition  is  thus  demonstrated. 

If  ABC,  DEF  be  two  triangles,  such  that  the  two  sides  AB  and  AC  of 
the  one  are  equal  to  the  two  ED,  DF  of  the  other,  and  the  angle  BAC, 
contained  by  the  sides  AB,  AC  of  the  one,  equal  to  the  angle  EDF,  con 
tained  by  the  sides  ED,  DF  of  the  other ;  the  triangles  ABC  and  EDF  are 
every  way  equal. 


NOTES. 


285 


On  AB  let  a  triangle  be  constituted  every  way  equal  to  the  triangle  DEF  ; 
then  if  this  triangle  coincide  with  the  triangle  ABC,  it  is  evident  that  the 
proposition  is  true,  for  it  is  equal  to  DEF  by  hypothesis,  and  to  ABC,  be- 
cause it  coincides  with  it ;  wherefore  ABC,  DEF  are  equal  to  one  another. 
But  if  it  does  not  coincide  with  ABC,  let  it  have  the  position  ABG ;  and  first 
suppose  G  not  to  fall  on  AC  ;  then  the  angle  BAG  is  not  equal  to  the  angle 
BAC.  But  the  angle  BAG  is  equal  to  the  angle  EDF,  therefore  EDF 
and  ABC  are  not  equal,  and  they  are  also  equal  by  hypothesis,  which  is 
impossible.  Therefore  the  point  G  must  fall  upon  AC  ;  now,  if  it  fall  upon 
AC  but  not  at  C,  then  AG  is  not  equal  to  AC  ;  but  AG  is  equal  to  DF, 
therefore  DF  and  AC  are  not  equal,  and  they  are  also  equal  by  supposition, 
which  is  impossible.  Therefore  G  must  coincide  with  C,  and  the  triangle 
AGB  with  the  triangle  ACB.  But  AGB  is  every  way  equal  to  DEF, 
therefore  ACB  and  DEF  are  also  every  way  equal. 

By  help  of  the  same  postulate,  the  fifth  may  also  be  very  easily  de- 
monstrated. 

Let  ABC  be  an  isosceles  triangle,  in  which  AB,  AC  are  the  equal  sides ; 
the  angle  ABC,  ACB  opposite  to  these  sides  are  also  equal. 

Draw  the  straight  line  EF  equal  to  BC,  and  suppose  that  on  EF  the  tri- 
angle DEF  is  constituted  every  way  equal  to  the  triangle  ABC,  that  is, 
having  DE  equal  to  AB,  DF  to  AC,  the  angle  EDF  to  the  angle  BAC,  the 
angle  ACB  to  the  angle  DFE,  &;c. 


Then  because  DE  is  equal  to  AB,  and  AB  is  equal  to  AC,  DE  is  equal 
to  AC  ;  and  for  the  same  reason,  DF  is  equal  to  AB.  And  because  DF  is 
equal  to  AB,  DE  to  AC,  and  the  angle  FDE  to  the  angle  BAC,  the  angle 
ABC  is  equal  to  the  angle  DFE.  But  the  angle  ACB  is  also,  by  hy- 
pothesis, equal  to  the  angle  DFE ;  therefore  the  angles  ABC,  ACB  are 
equal  to  one  another. 


286  NOTES. 

Such  demonstrations,  it  must,  however,  be  acknowledged,  trespass 
against  a  rule  which  Euclid  has  uniformly  adhered  to  throughout  the  Ele- 
ments, except  where  he  was  forced  by  necessity  to  depart  from  it ;  This 
rule  is,  that  nothing  is  ever  supposed  to  be  done,  the  manner  of  doing  which 
has  not  been  already  taught,  so  that  the  construction  is  derived  either  di- 
rectly from  the  three  postulates  laid  down  in  the  beginning,  or  from  pro- 
blems already  reduced  to  those  postulates.  Now,  this  rule  is  not  essential 
to  geometrical  demonstration,  where,  for  the  purpose  of  discovering  the 
properties  of  figures,  we  are  certainly  at  liberty  to  suppose  any  figure  to  be 
constructed,  or  any  line  to  be  drawn,  the  existence  of  which  does  not  in- 
volve an  impossibility.  The  only  use,  therefore,  of  Euclid's  rule  is  to 
guard  against  the  introduction  of  impossible  hypotheses,  or  the  taking  for 
granted  that  a  thing  may  exist  which  in  fact  implies  contradiction  ;  from 
such  suppositions,  false  conclusions  might,  no  doubt,  be  deduced,  and  the 
rule  is  therefore  useful  in  as  much  as  it  answers  the  purpose  of  excluding 
them.  But  the  foregoing  postulatum  could  never  lead  to  suppose  the 
actual  existence  of  any  thing  that  is  impossible ;  for  it  only  assumes  the 
existence  of  a  figure  equal  and  similar  to  one  already  existing,  but  in  a  dif- 
ferent part  of  space  from  it,  or  having  one  of  its  sides  in  an  assigned  posi- 
tion. As  there  is  no  impossibility  in  the  existence  oi  one  of  these  figures, 
it  is  evident  that  there  can  be  none  in  the  existence  of  the  other. 


PROP.  XXI.    THEOR. 

It  is  essential  to  the  truth  of  this  proposition,  that  the  straight  lines, 
drawn  to  the  point  within  the  triangle  be  drawn  from  the  two  extremities 
of  the  base  ;  for,  if  they  be  drawn  from  other  points  of  the  base,  their  sum 
may  exceed  the  sum  of  the  sides  of  the  triangle  in  any  ratio  that  is  less 
than  that  of  two  to  one.  This  is  demonstrated  by  Pappus  Alexandrinus 
in  the  3d  Book  of  his  Mathematical  CollectionSy  but  the  demonstration  is  of  a 
kind  that  does  not  belong  to  this  place.  If  it  be  required  simply  to  show> 
that  in  certain  cases  the  sum  of  the  two  lines  drawn  to  the  point  within  the 
triangle  may  exceed  the  sum  of  the  sides  of  the  triangle,  the  demonstra- 
tion is  easy,  and  is  given  nearly  as  follows  by  Pappus,  and  also  by  Proclus> 
in  the  4th  Book  of  his  Commentary  on  Euclid. 

Let  ABC  be  a  triangle,  having  the  angle  at  A  a  right  angle :  let  D  be 
any  point  in  AB  ;  join  CD,  then  CD  will  be  greater  than  AC,  because  in 
the  triangle  ACD  the  angle  CAD  is  greater  than  the  angle  ADC.  From 
DC  cut  oflf  DE  equal  to  AC  ;  bisect  CE 
in  F,  and  join  BF ;  BF  and  FD  are  greater 
than  BC  and  CA. 

Because  CF  is  equal  to  FE,  CF  and  FB 
are  equal  to  EF  and  FB,  but  CF  and  FB 
are  greater  than  BC,  therefore  EF  and  FB 
are  greater  than  BC.  To  EF  and  FB  add 
ED,  and  to  BC  add  AC,  which  is  equal  to 
ED  by  construction,  and  BF  and  FD  will 
be  greater  than  BC  and  CA. 


NOTES.  287 

It  is  evident,  that  if  the  angle  BAG  be  obtuse,  the  same  reasoning  may 
be  applied. 

This  proposition  is  a  sufficient  vindication  of  Euclid  for  having  demon- 
strated the  21st.  proposition,  w^hich  some  affect  to  consider  as  self-evident ; 
for  it  proves  that  the  circumstance  on  which  the  truth  of  that  proposition 
depends  is  not  obvious,  nor  that  which  at  first  sight  it  is  supposed  to  be,  viz. 
that  of  the  one  triangle  being  included  within  the  other.  For  this  reason  I 
cannot  agree  with  M.  Clairaut,  that  Euclid  domonstrated  this  proposition 
only  to  avoid  the  cavils  of  the  Sophists.  But  I  must,  at  the  same  time,  ob- 
serve, that  what  the  French  Geometer  has  said  on  the  subject  has  certain- 
ly been  misunderstood,  and  in  one  respect,  unjustly  censured  by  Dr.  Simson. 
The  exact  translation  of  his  words  is  as  follows  :  "  If  Euclid  has  taken  the 
"  trouble  to  demonstrate,  that  a  triangle  included  within  another  has  the 
"  sum  of  its  sides  less  than  the  sum  of  the  sides  of  the  triangle  in  which  it 
"is  included,  we  are  not  to  be  surprised.  That  Geometer  had  to  do  with 
"  those  obstinate  Sophists,  who  made  a  point  of  refusing  their  assent  to  the 
"  most  evident  truths,"  &c.  (Elements  de  Geometrie  par  M.  Clairaut. 
Pref.) 

Dr.  Simson  supposes  M.  Clairaut  to  mean,  by  the  proposition  which  he 
enunciates  here,  that  when  one  triangle  is  included  in  another,  the  sum  of 
the  two  sides  of  the  included  triangle  is  necessarily  less  than  the  sum  of  the 
two  sides  of  the  triangle  in  which  it  is  included,  whether  they  be  on  the 
same  base  or  not.  Now  this  is  not  only  not  Euclid's  proposition,  as  Dr 
Simson  remarks,  but  it  is  not  true,  and  is  directly  contrary  to  what  has 
just  been  demonstrated  from  Proclus.  But  the  fact  seems  to  be,  that  M. 
Clairaut's  meaning  is  entirely  different,  and  that  he  intends  to  speak  not  of 
two  of  the  sides  of  a  triangle,  but  of  all  the  three  ;  so  that  his  proposition 
is,  "  that  when  one  triangle  is  included  within  another,  the  sum  of  all  the 
"three  sides  of  the  included  triangle  is  less  than  the  sum  of  all  the  three 
"  sides  of  the  other,"  and  this  is  without  doubt  true,  though  I  think  by  no 
means  self-evident.  It  must  be  acknowledged  also,  that  it  is  not  exactly 
Euclid's  proposition,  which,  however,  it  comprehends  under  it,  and  is  the 
general  theorem,  of  which  the  other  is  only  a  particular  case.  Therefore, 
though  M.  Clairaut  may  be  blamed  for  maintaining  that  to  be  an  Axiom 
which  requires  demonstration,  yet  he  is  not  to  be  accused  of  mistaking  a 
false  proposition  for  a  true  one. 

PROP.  XXII.    PROB. 

Thomas  Simson  in  his  Elements  has  objected  to  Euclid's  demonstration 
of  this  proposition,  because  it  contains  no  proof,  that  the  two  circles  made 
use  of  in  the  construction  of  the  Problem  must  cut  one  another ;  and  Dr. 
Simson  on  the  other  hand,  always  unwilling  to  acknowledge  the  smallest 
blemish  in  the  works  of  Euclid,  contends  that  the  demonstration  is  perfect. 
The  truth,  however,  certainly  is,  that  the  demonstration  admits  of  some 
improvement ;  for  the  limitation  that  is  made  in  the  enunciation  of  any 
Problem  ought  always  to  be  shewn  to  be  necessarily  connected  with  the 
construction  of  it,  and  this  is  what  Euclid  has  neglected  to  do  in  the  pre- 
sent instance.     The  defect  may  easily  be  supplied,  and  Dr.  Simson  him- 


288  NOTES. 

self  has  done  it  in  effect  in  his  note  on  this  proposition,  though  he  denies  it 
to  be  necessary. 

Because  that  of  the  three  straight  lines  DF,  FG,  GH,  any  two  are  great- 
er than  the  third,  by  hypothesis,  FD  is  less  than  FG  and  GH,  that  is, 
than  FH,  and  therefore  the  circle  described  from  the  centre  F,  with  the 
distance  FD  must  meet  the  line  FE  between  F  and  H  ;  and,  for  the  like 


reason,  the  circle  described  from  the  centre  G  at  the  distance  GH,  must 
meet  DG  between  D  and  G,  and  therefore  the  one  of  thes©  circles  can- 
not be  wholly  within  the  other.  Neither  can  the  one  be  wholly  without 
the  other,  because  DF  and  GH  are  greater  than  FG ;  the  two  circles 
must  therefore  intersect  one  another. 

PROP.  XXVn.  and  XXVHI. 

Euclid  has  been  guilty  of  a  slight  inaccuracy  in  the  enunciations  of 
these  propositions,  by  omitting  the  condition,  that  the  two  straight  lines  on 
which  the  third  line  falls,  making  the  alternate  angles,  &c.  equal,  must 
be  in  the  same  plane,  without  which  they  cannot  be  parallel,  as  is  evident 
from  the  definition  of  parallel  lines.  The  only  editor,  I  believe,  who  has  re- 
marked this  omission,  is  M.  de  Foix  Due  de  Candalle,  in  his  transla- 
tion of  the  Elements  published  in  1566.  How  it  has  escaped  the  notice  of 
subsequent  commentators  is  not  easily  explained,  unless  because  they 
thought  it  of  little  importance  to  correct  an  error  by  which  nobody  was 
likely  to  be  misled. 

PROP.  XXIX. 

The  subject  of  parallel  lines  is  one  of  the  most  difficult  in  the  Elements 
of  Geometry.  It  has  accordingly  been  treated  of  in  a  great  variety  of  differ- 
ent ways,  of  which,  perhaps,  there  is  none  that  can  be  said  to  have  given 
entire  satisfaction.  The  difficulty  consists  in  converting  the  27th  and  28th  of 
Euclid,  or  in  demonstrating,  that  parallel  straight  lines,  or  such  as  do  not 
meet  one  another,  when  they  meet  a  third  line,  make  the  alternate  angles 
with  it  equal,  or,  which  comes  to  the  same,  are  equally  inclined  to  it,  and 
make  the  exterior  angle  equal  to  the  interior  and  opposite.     In  order  to  de- 


NOTES.  289 

monstrate  this  proposition,  Euclid  assumed  it  as  an  Axiom,  that  "  if  a 
"  straight  line  meet  two  straight  lines,  so  as  to  make  the  interior  angles  on 
"  the  same  side  of  it  less  than  two  right  angles,  these  straight  lines  being 
"  continually  produced,  will  at  length  meet  on  the  side  on  which  the  angles 
"  are  that  are  less  than  two  right  angles."  This  proposition,  however,  is 
not  self-evident,  and  ought  the  less  to  be  received  without  proof,  that,  as 
Proclus  has  observed,  the  converse  of  it  is  a  proposition  that  confessedly 
requires  to  -be  demonstrated.  For  the  converse  of  it  is,  that  two  straight 
lines  which  meet  one  another  make  the  interior  angles,  with  any  third  line, 
less  than  two  right  angles  ;  or,  in  other  words,  that  the  two  interior  angles 
of  any  triangle  are  less  than  two  right  angles,  which  is  the  17th  of  the 
First  Book  of  the  Elements :  and  it  should  seem,  that  a  proposition  can 
never  rightly  be  taken  for  an  Axiom,  of  which  the  converse  requires  a  de- 
monstration. 

The  methods  by  which  Geometers  have  attempted  to  remove  this 
blemish  from  the  Elements  are  of  three  kinds.  1 .  By  a  new  definition  of 
parallel  lines.  2.  By  introducing  a  new  Axiom  concerning  parallel  lines, 
more  obvious  than  Euclid's.  3.  By  reasoning  merely  from  the  definition 
of  parallels,  and  the  properties  of  lines  already  demonstrated  without  the 
assumption  of  any  new  Axiom. 

1 .  One  of  the  definitions  that  has  been  substituted  for  Euclid's  is,  that 
straight  lines  are  parallel,  which  preserve  always  the  same  distance  from 
one  another,  by  the  word  distance  being  understood,  a  perpendicular  drawn 
to  one  of  the  lines  from  any  point  whatever  in  the  other.  If  these  perpendicu- 
lars be  every  where  of  the  same  length,  the  straight  lines  are  called  parallel. 
This  is  the  definition  given  by  Woltius,  by  Boscovich,  and  by  Thomas 
Simson,  in  the  first  edition  of  his  Elements.  It  is  however  a  faulty  defi- 
nition, for  it  conceals  an  Axiom  in  it,  and  takes  for  granted  a  property  of 
straight  lines,  that  ought  either  to  be  laid  down  as  self-evident,  or  demonstrat- 
ed, impossible,  as  a  Theorem.  Thus,  if  from  the  three  points,  A,  B,  and  C 
of  the  straight  line  AC,  perpendiculars  AD,  BE,  CF  be  drawn  all  equal 
to  one  another,  it  is  implied  in  the  definition 
that  the  points  D,  E  and  F  are  in  the  same 
straight  line,  which,  though  it  be  true,  it  was 
not  the  business  of  the  definition  to  inform  us 
of.  Two  perpendiculars,  as  AD  and  CF,  are 
alone  sufficient  to  determine  the  position  of  the 
straight  line  DF,  and  therefore  the  definition  ought  to  be,  "  that  two  straight 
"  lines  are  parallel,  when  there  are  two  points  in  the  one,  from  which  the 
"  perpendiculars  drawn  to  the  other  are  equal,  and  on  the  same  side  of  it." 

This  is  the  definition  of  parallels  which  M.  D'Alembert  seems  to  prefer 
to  all  others  ;  but  he  acknowledges,  and  very  justly,  that  it  still  remains  a 
matter  of  difficulty  to  demonstrate,  that  all  the  perpendiculars  drawn  from 
the  one  of  these  lines  to  the  other  are  equal.     {Encyclopedic,  Art,  Parallele.) 

Another  definition  that  has  been  given  of  parallels  is,  that  they  are  lines 
which  make  equal  angles  with  a  third  line,  toward  the  same  parts,  or  such 
as  make  the  exterior  angle  equal  to  the  interior  and  opposite.  Varignon, 
Bezout,  and  several  other  mathematicians,  have  adopted  this  definition, 
which,  it  must  be  acknowledged,  is  a  perfectly  good  one,  if  it  be  understood 

37 


A. 


390  NOTES. 

by  it,  that  the  two  lines  called  parallel,  are  such  as  make  equal  angles  with 

A.    \g  B 


a  certain  third  line,  but  not  with  any  line  that  falls  upon  them.  It  remains, 
therefore,  to  be  demonstrated,  That  if  AB  and  CD  make  equal  angles  with 
GH,  they  will  do  so  also  with  any  other  line  whatsoever.  The  definition, 
therefore,  must  be  thus  understood,  That  parallel  lines  are  such  as  make 
equal  angles,  with  a  certain  third  line,  or,  more  simply,  lines  which  are  per- 
pendicular to  a  given  line.  It  must  then  be  proved,  1.  That  straight  lines 
which  are  equally  inclined  to  a  certainlinc  or  perpendicular  to  a  certain  line, 
must  be  equally  inclined  to  all  the  other  lines  that  fall  upon  them ;  and  also, 
2.  That  two  straight  lines  which  do  not  meet  when  produced,  must  make 
equal  angles  with  any  third  line  that  meets  them. 

The  demonstration  of  the  first  of  these  propositions  is  not  at  all  facilitated 
by  the  new  definition,  unless  it  be  previously  shown  that  all  the  angles  of  a 
triangle  are  equal  to  two  right  angles. 

The  second  proposition  would  hardly  be  necessary  if  the  new  definition 
were  employed  ;  for  when  it  is  required  to  draw  a  line  that  shall  not  meet 
a  given  line,  this  is  done  by  drawing  a  line  that  shall  have  the  same  incli- 
nation to  a  third  line  that  the  first  or  given  line  has.  It  is  known  that  lines 
so  drawn  cannot  meet.  It  would  no  doubt  be  an  advantage  to  have  a  defi- 
nition that  is  not  founded  on  a  condition  purely  negative. 

2.  As  to  the  Mathematicians  who  have  rejected  Euclid's  Axiom,  and  in- 
troduced another  in  its  place,  it  is  not  necessary  that  much  should  be  said. 
Clavius  is  one  of  the  first  in  this  class  ;  the  Axiom  he  assumes  is,  "  That  a 
"  line  of  which  the  points  are  all  equidistant  from  a  certain  straight  line  in 
"  the  same  plane  with  it,  is  itself  a  straight  line."  This  proposition  he  does 
not,  however,  assume  altogether,  as  he  gives  a  kind  of  metaphysical  proof 
of  it,  by  which  he  endeavours  to  connect  it  with  Euclid's  definition  of  a 
straight  line,  with  which  proof  at  the  same  time  he  seems  not  very  well 
satisfied.  His  reasoning,  after  this  proposition  is  granted  (though  it  ought 
not  to  be  granted  as  an  Axiom),  is  logical  and  conclusive,  but  is  prolix  and 
operose,  so  as  to  leave  a  strong  suspicion  that  the  road  pursued  is  by  no 
means  the  shortest  possible. 

The  method  pursued  by  Simson,  in  his  Notes  in  the  First  Book  of  Euclid, 
is  not  very  different  from  that  of  Clavius.  He  assumes  this  Axiom, "  That 
"  a  straight  line  cannot  first  come  nearer  to  another  straight  line,  and  then 
"  go  farther  from  it  without  meeting  it."  (Notes,  (fee.  English  Edition.)  By 
coming  nearer  is  understood,  conformably  to  a  previous  definition,  the  dirai- 


NOTES. 


nulion  of  the  perpendiculars  drawn  from  the  one  line  to  the  other.  This 
Axiom  is  more  readily  assented  to  than  that  of  Clavius,  from  which,  how- 
ever, it  is  not  very  different :  but  it  is  not  very  happily  expressed,  as  the  idea 
not  merely  of  motion,  but  of  time,  seems  to  be  involved  in  the  notion  o{ first 
coming  nearer,  and  then  going  farther  off.  Even  if  this  inaccuracy  is  pass- 
ed over,  the  reasoning  of  Simson,  like  that  of  Clavius,  is  prolix,  and  evi- 
dently a  circuitous  mei,hod  of  coming  at  the  truth. 

Thomas  Simson,  in  the  second  edition  of  his  Elements,  has  presented 
this  Axiom  in  a  simpler  form.  "  If  two  points  in  a  straight  line  are  posited 
"at  unequal  distances  from  another  straight  line  in  the  same  plane, 
"  those  two  lines  being  indefinitely  produced  on  the  side  of  the  least  dis- 
"  tance  will  meet  one  another." 

By  help  of  this  Axiom  it  is  easy  to  prove,  that  if  two  straight  lines  AB, 
CD  are  parallel,  the  perpendiculars  to  the  one,  terminated  by  the  other, 
are  all  equal,  and  are  also  perpendicular  to  both  the  parallels.  That  they 
are  equal  is  evident,  otherwise  the  lines  would  meet  by  the  Axiom.  That 
they  are  perpendicular  to  both,  is  demonstrated  thus  : 

If  AC  and  BD,which  are  perpendicular  to  AB,  and  equal  to  one  another, 
be  not  also  perpendicular  to  CD,  fr9m  C  let  CE 
be  drawn  at  right  angles  to  BD.  Then,  be- 
cause AB  and  CE  are  both  perpendicular  to 
BD,  they  are  parallel,  and  therefore  the  perpen- 
diculars AC  and  BE  are  equal.  But  AC  is 
equal  to  BD,  (by  hypotheses,)  therefore  BE  and  J^ 
BD  are  equal,  which  is  impossible  ;  BD  is  therefore  at  right  angles  to  CD. 

Hence  the  proposition,  that  "  if  a  straight  line  fall  on  two  parallel  lines,  it 
"makes  the  alternate  angles  equal,"  is  easily  derived.    Let  FH  and  GE  be 


perpendicular  to  CD,  then  they  will  be  parallel  to  one  another,  and  also  at 
right  angles  to  AB,  and  therefore  FG  and  HE  are  equal  to  one  another, 
by  the  last  proposition.  Wherefore  in  the  triangles  EFG,  EFH,  the  sides 
HE  and  EF  are  equal  to  the  sides  GF  and  FE,  each  to  each,  and  also  the 
third  side  HF  to  the  third  side  EG,  therefore  the  angle  HEF  is  equal  to 
the  angle  EFG,  and  they  are  alternate  angles. 

This  method  of  treating  the  doctrine  of  parallel  lines  is  extremely  plain 
and  concise,  and  is  perhaps  as  good  as  any  that  can  be  followed,  when  a 
new  Axiom  is  assumed.  In  the  text  above,  I  have,  however,  followed  a 
different  method,  employing  as  an  Axiom,  "That  two  straight  lines,  which 
"  cut  one  another,  cannot  be  both  parallel  to  the  same  straight  line."  This 
Axiom  has  been  assumed  by  others,  particularly  by  Ludlam,  in  his  very 
useful  little  tract,  entitled  Rudiments  of  Mathematics. 


293  NOTES.  '  * 

It  is  a  proposition  readily  enough  admitted  as  self-evident,  and  leads 
to  the  demonstration  of  Euclid's  29th  Proposition,  even  with  more  brevity 
than  Simson's. 

3.  All  the  methods  above  enumerated  leave  the  mind  somewhat  dissatis- 
fied, as  we  naturally  expect  to  discover  the  properties  of  parallel  lines,  as 
we  do  those  of  other  geometric  quantities,  by  comparing  the  definition  of 
those  lines,  with  the  properties  of  straight  lines  alrea^dy  known.  The  most 
ancient  writer  who  appears  to  have  attempted  to  do  this  is  Ptolemy  the  as- 
tronomer, who  wrote  a  treatise  expressly  on  the  subject  of  Parallel  Lines. 
Proclus  has  preserved  some  account  of  this  work  in  the  Fourth  Book  of  his 
commentaries  :  and  it  is  curious  to  observe  in  it  an  argument  founded  on  the 
principle  which  is  known  to  the  modems  by  the  name  of  the  sufficient  reasm. 

To  prove,  that  if  two  parallel  straight  lines,  AB  and  CD,  be  cut  by  a 
third  line  EF,  in  G  and  H,  the  two  interior  angles  AGH,  GHG  will  be 


B 

zv 

C               H\ 

equal  to  two  right  angles,  Ptolemy  reasons  thus :  If  the  angles  AGH, 
CHG  be  not  equal  to  two  right  angles,  let  them,  if  possible,  be  greater 
than  two  right  angles :  then,  because  the  lines  AG  and  CH  are  not  more 
parallel  than  the  lines  BG  and  DH,  the  angles  BGH,  DHG  are  also 
greater  than  two  right  angles.  Therefore,  the  four  angles  AGH,  CHG, 
BGH,  DHG  are  greater  than  four  right  angles  ;  and  they  are  also  equal 
to  four  right  angles,  which  is  absurd.  In  the  same  manner  it  is  shewn, 
that  the  angles  AGH,  CHG  cannot  be  less  than  two  right  angles.  There- 
fore they  are  equal  to  tv/o  right  angles. 

But  this  reasoning  is  certainly  inconclusive.  For  why  are  we  to  sup- 
pose that  the  interior  angles  which  the  parallels  make  wiih  the  line  cutting 
them,  are  either  in  every  case  greater  than  two  right  angles,  or  in  every 
case  less  than  two  right  angles  1  For  any  thing  that  we  are  yet  supposed 
to  know,  they  may  be  sometimes  greater  than  two  right  angles,  and  some- 
times less,  and  therefore  we  are  not  entitled  to  conclude,  because  the  angles 
AGH,  CHG  are  greater  than  two  right  angles,  that  therefore  the  angles 
BGH,  DHG  are  also  necessarily  greater  than  two  right  angles.  It 
may  safely  be  asserted,  therefore,  that  Ptolemy  has  not  succeeded  in  his 
attempt  to  demonstrate  the  properties  of  parallel  lines  without  the  assist- 
ance of  a  new  Axiom. 

Another  attempt  to  demonstrate  the  same  proposition  without  the  assist- 
ance of  a  new  Axiom  has  been  made  by  a  modern  geometer,  Franceschini, 


NOTES. 


293 


:^  N 


Professor  of  Mathematics  in  the  University  of  Bologna,  in  an  essay,  which 
he  entitles,  La  Teoria  dclle  parallele  rigor osamente  dimonstrata,  printed  in 
his  Opuscoli  Mathematici,  at  Bassano  in  1787. 

The  difficulty  is  there  reduced  to  a  proposition  nearly  the  same  with  this, 
That  if  BE  make  an  acute  angle  with  BD,  and  if  DE  be  perpendicular  to 
BD  at  any  point,  BE  and  DE, 
if  produced,  will  meet.  To  de- 
monstrate this,  it  is  supposed, 
that  BO,  BC  are  two  parts  taken 
in  BE,  of  which  BG  is  greater 
than  BO,  and  that  the  perpendi- 
culars ON,  CL  are  drawn  to  BD  ; 
then  shall  BL  be  greater  than 
BN.  For,  if  not,  that  is,  if  the 
perpendicular  CL  falls  either  at 
N,  or  between  B  and  N,  as  at 
F ;  in  the  first  of  these  cases  the 
angle  CNB  is  equal  to  the  angle  ONB,  because  they  are  both  right  angles, 
which  is  impossible  ;  and,  in  the  second,  the  two  angles  CFN,  CNF  of  the 
triangle  CNF,  exceed  two- right  angles.  Therefore,  adds  our  author,  since, 
as  BC  increases,  BL  also  increases,  and  since  BC  may  be  increased  with- 
out limit,  so  BL  may  become  greater  than  any  given  line,  and  therefore  may 
be  greater  than  BD  ;  wherefore,  since  the  perpendiculars  to  BD  from  points 
beyond  D  meet  BC,  the  perpendicular  from  D  necessarily  meets  it. 

Now  it  will  be  found,  on  examination,  that  this  reasoning  is  no  more 
conclusive  than  the  preceding.  For,  unless  it  be  proved,  that  whatever 
multiple  BC  is  of  BO,  the  same  is  BL  of  BN,  the  indefinite  increase  of 
BC  does  not  necessarily  imply  the  indefinite  increase  of  BL,or  that  BL  may 
be  made  to  exceed  BD.  On  the  contrary,  BL  may  always  increase,  and 
yet  may  do  so  in  such  a  manner  as  never  to  exceed  BD  :  In  order  that  the 
demonstration  should  be  conclusive,  it  would  be  necessary  to  shew,  that 
when  BC  increases  by  a  part  equal  to  BO,  BL  increases  always  by  a  part 
equal  to  BN  ;  but  to  do  this  will  be  found  to  require  the  knowledge  of  those 
very  properties  of  parallel  lines  that  we  are  seeking  to  demonstrate. 

Legendre,  in  his  Elements  of  Geometry,  a  work  entitled  to  the  highest 
praise,  for  elegance  and  accuracy,  has  delivered  the  doctrine  of  parallel  lines 
without  any  new  Axiom.  He  has  done  this  in  two  dilTerent  ways,  one  in 
the  text,  and  the  other  in  the  notes.  In  the  former  he  has  endeavoured  to 
prove,  independently  of  the  doctrine  of  parallel  lines,  that  all  the  angles  of 
a  triangle  are  equal  .to  two  right  angles  ;  from  which  proposition,  when 
it  is  once  established,  it  is  not  difficult  to  deduce  every  thing  with  respect  to 
parallels.  But,  though  his  demonstration  of  the  property  of  triangles  just 
mentioned  is  quite  logical  and  conclusive,  yet  it  has  the  fault  of  being  long 
and  indirect,  proving  first,  that  the  three  angles  of  a  triangle  cannot  be 
greater  than  two  right  angles,  next,  that  they  cannot  be  less,  and  doing 
both  by  reasoning  abundantly  subtle,  and  not  of  a  kind  readily  apprehend- 
ed by  those  who  are  only  beginning  to  study  the  Mathematics. 

The  demonstration  which  he  has  given  in  the  notes  is  extremely  ingeni- 
ous, and  proceeds  on  this  very  simple  and  undeniable  Axiom,  that  we  can- 
not compare  an  angle  and  a  line,  as  to  magnitude,  or  cannot  have  an  equa- 


294  NOTES,  / 

tion  of  any  sort  between  them.  This  truth  is  involved  in  the  distinction 
between  homogeneous  and  heterogeneous  quantities,  (Euc.  v.  def.  4.), 
which  has  long  been  received  in  Geometry,  but  led  only  to  negative  con- 
sequences, till  it  fell  into  the  hands  of  Legendre.  The  proposition  which 
he  deduces  from  it  is,  that  if  two  angles  of  one  triangle  be  equal  to  two  an- 
gles of  another,  the  third  angles  of  these  triangles  are  also  equal.  For,  it 
is  evident,that  when  two  angles  of  a  triangle  are  given,  and  also  the  side 
between  them,  the  third  angle  is  thereby  determined  ;  so  that  if  A  and  B 
be  any  two  angles  of  a  triangle,  P  the  side  interjacent,  and  C  the  third  an- 
gle, C  is  determined,  as  to  its  magnitude,  by  A,  B  and  P ;  and,  besides 
these,  there  is  no  other  quantity  whatever  which  can  afTect  the  magnitude 
of  C.  This  is  plain,  because  if  A,  B  and  P  are  given,  the  triangle  can  be 
constructed,  all  the  triangles  in  which  A,  B  and  P  are  the  same,  being  equal 
to  one  another. 

But  of  the  quantities  by  which  C  is  determined,  P  cannot  be  one  ;  for  if 
it  were,  then  C  must  be  b.  function  of  the  quantities  A,  B,  P  ;  that  is  to  say, 
the  value  of  C  can  be  expressed  by  some  combination  of  the  quantities  A, 
B  and  P.  An  equation,  therefore,  may  exist  between  the  quantities  A,  B 
C  and  P ;  and  consequently  the  value  of  P  is  equal  to  some  combination 
that  is,  to  some  fimction  of  the  quantities  A,  B  and  C  ;  but  this  is  impossi- 
ble, P  being  a  line,  and  A,  B,  C  being  angles  ;  so  that  no  function  of  the 
first  of  these  quantities  can  be  equal  to  any  function  of  the  other  three.  The 
angle  C  must  therefore  be  determined  by  the  angles  A  and  B  alone,  without 
any  regard  to  the  magnitude  of  P,  the  side  interjacent.  Hence  in  all  trian- 
gles that  have  two  angles  in  one  equal  to  two  in  another,  each  to  each  the 
third  angles  are  also  equal. 

Now,  this  being  demonstrated,  it  is  easy  to  prove  that  the  three  angles  of 
any  triangle  are  equal  to  two  right  angles. 

Let  ABC  be  a  triangle  right  angled  at  A,  draw  AD  perpendicular  to 
BC.     The  triangles  ABD,  ABC  have  the  an-  a 

gles  BAC,  BDA  right  angles,  and  the  angle 
B  common  to  both ;  therefore  by  what  has  just 
been  proved,  their  third  angles  BAD,  BCA  are 
also  equal.  In  the  same  way  it  is  shewn,  that 
CAD  is  equal  to  CBA  ;  therefore  the  two  an- 
gles, BAD,  CAD  are  equal  to  the  two  BCA, 
CBA;  but  BAD-fCAD  is  equal  to  a  right  B 
angle,  therefore  the  angles  BCA,  CBA  are  together  equal  to  a  right  angle, 
and  consequently  the  three  angles  of  the  right  angled  triangle  ABC  are 
equal  to  two  right  angles. 

And  since  it  is  proved  that  the  oblique  angles  of  every  right  angled 
triangle  are  equal  to  one  right  angle,  and  since  every  triangle  may  be 
divided  into  two  right  angled  triangles,  the  four  oblique  angles  of  which  are 
equal  to  the  three  angles  of  the  triangle,  therefore  the  three  angles  of  every 
triangle  are  equal  to  two  right  angles. 

Though  this  method  of  treating  the  subject  is  strictly  demonstrative,  yet, 
as  the  reasoning  in  the  first  of  the  two  preceding  demonstrations  is  not  per- 
haps sufficiently  simple  to  be  apprehended  by  those  just  entering  on  mathe- 
matical studies,  I  shall  submit  to  the  reader  another  method,  not  liable  to 
the  same  objection,  which  I  know,  from  experience,  to  be  of  use  in  explain- 


NOTES.  293 

ing  tlie  Elements.  It  proceeds,  like  that  of  the  French  Geometer,  by  de- 
monstrating, in  the  first  place,  that  the  angles  of  any  triangle  are  together 
equal  to  two  right  angles,  and  deducing  from  thence,  that  two  lines,  which 
make  with  a  third  line  the  interior  angles,  less  than  two  right  angles,  must 
meet  if  produced.  The  reasoning  used  to  demonstrate  the  first  of  these 
propositions  may  be  objected  to  by  some  as  involving  the  idea  of  motion,  and 
the  transference  of  a  line  from  one  place  to  another.  This,  however,  is  no 
more  than  Euclid  has  done  himself  on  some  occasions ;  and  when  it  furnish- 
es so  short  a  road  to  the  truth  as  in  the  present  instance,  and  does  not  im- 
pair the  evidence  of  the  conclusion,  it  seems  to  be  in  no  respect  inconsistent 
with  the  utmost  rigour  of  demonstration.  It  is  of  importance  in  explaining 
the  Elements  of  Science,  to  connect  truths  by  the  shortest  chain  possible  ; 
and  till  that  is  done,  we  can  never  consider  them  as  being  placed  in  their 
natural  order.  The  reasoning  in  the  first  of  the  following  propositions  is  so 
simple,  that  it  seems  hardly  susceptible  of  abbreviation,  and  it  has  the  ad- 
vantage of  connecting  immediately  two  truths  so  much  alike,  that  one 
might  conclude,  even  from  the  bare  enunciations,  that  they  are  but  different 
cases  of  the  same  general  theorem,  viz.  That  all  the  angles  about  a  point, 
and  all  the  exterior  angles  of  any  rectilineal  figure,  are  constantly  of  the 
same  magnitude,  and  equal  to  four  right  angles. 

DEFINITION. 

If,  while  one  extremity  of  a  straight  line  re- 
mains fixed  at  A,  the  line  itself  turns  about  that 
point  from  the  position  AB  to  the  position  AC,  it 
is  said  to  describe  the  angle  BAG  contained  by 
the  line  AB  and  AG. 

GoR.  If  a  line  turn  about  a  point  from  the  position  AG  till  it  come  into 
the  position  AG  again,  it  describes  angles  which  are  together  equal  to  four 
right  angles.     This  is  evident  from  the  second  Gor.  to  the  15th.  1. 

PROP.  I. 

All  the  exterior  angles  of  any  rectilineal  figure  are  together  equal  to  four 
right  angles. 

1.  Let  the  rectilineal  figure  be  the  triangle  ABG,  ^f  which  the  exterior 
angles  are  DGA,  FAB,  GBG ;  these  angles  are  together  equal  to  four 
right  angles. 

Let  the  line  GD,  placed  in  the  direction  of  BG  produced,  turn  about  the 
point  G  till  it  coincide  with  GE,  a  part  of  the  side  GA,  and  have  described 
the  exterior  angle  DGE  or  DGA.  Let  it  then  be  carried  along  the  line 
GA,  till  it  be  in  the  position  AF,  that  is,  in  the  direction  of  GA  produced, 
and  ihe  point  A  remaining  fixed,  let  it  turn  about  A  till  it  describe  the 
angle  FAB,  and  coincide  with  a  part  of  the  line  AB.  Let  it  next  be  car- 
ried  along  AB  till  it  come  into  the  position  BG,  and  by  turning  about  B, 


296 


NOTES. 


let  it  describe  the  angle  GBC,  so 
as  to  coincide  with  a  part  of  BC. 
Lastly,  Let  it  be  carried  along  BC 
till  it  coincide  with  CD,  its  first 
position.  Then,  because  the  line 
CD  has  turned  about  one  of  its 
extremities  till  it  has  come  into 
the  position  CD  again,  it  has  by 
the  corollary  to  the  above  defini- 
tion described  angles  which  are 
together  equal  to  four  right  an- 
gles ;  but  the  angles  which  it 
has  described  are  the  three  ex- 
terior angles  of  the  triangle  ABC, 
therefore  the  exterior  angles  of 
the  triangle  ABC  are  equal  to 
four  right  angles. 

2.  If  the  rectilineal  figure  have  any  number  of  sides,  the  proposition  is 
demonstrated  just  as  in  the  case  of  a  triangle.  Therefore  all  the  exterior 
angles  of  any  rectilineal  figure  are  together  equal  to  four  right  angles. 

CoR.  1.  Hence,  all  the  interior  angles  of  any  triangle  are  equal  to  two 
right  angles.  For  all  the  angles  of  the  triangle,  both  exterior  and  interior, 
are  equal  to  six  right  angles,  and  the  exterior  being  equal  to  four  right 
angles,  the  interior  are  equal  to  two  right  angles. 

Cor.  2.  An  exterior  angle  of  any  triangle  is  equal  to  the  two  interior  and 
opposite,  or  the  angle  DC  A  is  equal  to  the  angles  CAB,  ABC.  For  the 
angles  CAB,  ABC,  BCA  are  equal  to  two  right  angles ;  and  the  angles 
ACD,  ACB  are  also  (13. 1.)  equal  to  two  right  angles  ;  therefore  the  three 
angles  CAB,  ABC,  BCA  are  equal  to  the  two  ACD,  ACB  ;  and  taking 
ACB  from  both,  the  angle  ACD  is  equal  to  the  two  angles  CAB,  ABC. 

CoR.  3.  The  interior  angles  of  any  rectilineal  figure  are  equal  to  twice 
as  many  right  angles  as  the  figure  has  sides,  wanting  four.  For  all  the 
angles  exterior  and  interior  are  equal  to  twice  as  many  right  angles  as  the 
figure  has  sides  ;  but  the  exterior  are  equal  to  four  right  angles  ;  therefore 
the  interior  are  equal  to  twice  as  many  right  angles  as  the  figure  has  sides, 
wanting  four. 

PROP.  IL 


Two  straight  lines,  which  make  with  a  third  line  the  interior  angles  on 
the  same  side  of  it  less  than  two  right  angles,  will  meet  on  that  side,  if  pro- 
duced far  enough. 

Let  the  straight  lines  AB,  CD,  make  with  AC  the  two  angles  BAC, 
DC  A  less  than  two  right  angles ;  AB  and  CD  will  meet  if  produced  toward 
B  and  D. 

In  AB  take  AF=AC  ;  join  CF,  produce  BA  to  H,  and  through  C  draw 
CE,  making  the  angle  ACE  equal  to  the  angle  CAH. 

Because  AC  is  equal  to  AF,  the  angles  AFC,  ACF  aie  also  equal  (5. 


NOTES. 


297 


1.) ;  but  the  exterior  angle  HAC  is  equal  to  the  two  interior  and  opposite 
angles  ACF,  AFC,  and  therefore  it  is  double  of  either  of  them,  as  of  ACF. 
Now  ACE  is  equal  to  HAC  by  construction,  therefore  ACE  is  double  of 
ACF,  and  is  bisected  by  the  line  CF.  In  the  same  manner,  if  FG  be  taken 
equal' to  FC,  and  if  CG  be  drawn,  it  may  be  shewn  that  CG  bisects  the 
angle  FCE,  and  so  on  continually.  But  if  from  a  magnitude,  as  the  an- 
gle ACE,  there  be  taken  its  half,  and  from  the  remainder  FCE  its 
half  FCG,  and  from  the  remainder  GCE  its  half,  &c.  a  remainder  will  at 
length  be  found  less  than  the  given  angle  DCE.* 


K      A 


Let  GCE  be  the  angle,  whose  half  ECK  is  less  than  DCE,  then  a 
straight  line  CK  is  found,  which  falls  between  CD  and  CE,  but  never- 
theless meets  the  line  AB  in  K.  Therefore  CD,  if  produced,  must  meet 
AB  in  a  point  between  G  and  K. 

This  demonstration  is  indirect ;  but  this  proposition,  if  the  definition  of 
parallels  were  changed,  as  suggested  at  p.  291,  would  not  be  necessary  ; 
and  the  proof,  that  lines  equally  inclined  to  any  one  line  must  be  so  to 
every  line,  would  follow  directly  from  the  angles  of  a  triangle  being  equal 
to  two  right  angles.  The  doctrine  of  parallel  lines  would  in  this  manner 
be  freed  from  all  difficulty. 

PROP.  III.  or  29.  I.Euclid. 


If  a  straight  line  fall  on  two  parallel  straight  lines,  it  makes  the  alternate 
angles  equal  to  one  another  ;  the  exterior  equal  to  the  interior  and  oppo- 
site on  the  same  side  ;  and  likewise  the  two  interior  angles,  on  the  same 
side  equal  to  two  right  angles. 

Let  the  straight  line  EF  fall  on 
the  parallel  straight  lines  AB, 
CD  ;  the  alternate  angles  AGH, 
GHD  are  equal,  the  exterior  angle 
EGB  is  equal  to  the  interior  and 
opposite  GHD  ;  and  the  two  inte- 
rior angles  BGH,  GHD  are  equal 
to  two  right  angles. 

For  if  AGH  be  not  equal  to 
GHD,  let  it  be  greater,  then  add- 
ing BGH  to  both,  the  angles 
AGH,  HGB  are  greater  than  the 


*  Prop.  1.  1  Sup.  The  reference  of  this  proposition  involves  nothing  inccns'stent  with 
good  reasoning,  as  the  demonstration  of  it  does  not  depend  on  any  thing  that  has  gone  before, 
so  that  it  msy  be  introduced  in  any  part  of  the  Elements. 

38 


298  NOTES. 

angles  DHG,  HGB.  But  AGH,  H6B  are  equal  to  two  right  angles  (13. 
1.) ;  therefore  BGH,  GHD  are  less  than  two  right  angles,  and  therefore  the 
lines  AB,  CD  will  meet,  by  the  last  proposition,  if  produced  toward  B  and 
D.  But  they  do  not  meet,  for  they  are  parallel  by  hypotheses,  and  there- 
fore the  angles  AGH,  GHD  are  not  unequal,  that  is,  they  are  equal  to  one 
another. 

Now  the  angle  AGH  is  equal  to  EGB,  because  these  are  vertical,  and 
it  has  also  been  shewn  to  be  equal  to  GHD,  therefore  EGB  and  GHD  are 
equal.  Lastly,  to  each  of  the  equal  angles  EGB,  GHD  add  the  angle 
BGH,  then  the  two  EGB,  BGH  are  equal  to  the  two  DHG,  BGH.  But 
EGB,  BGH  are  equal  to  two  right  angles  (13.  l.),therefore  BGH,  GHD 
are  also  equal  to  two  right  angles. 


The  following  proposition  is  placed  here,  because  it  is  more  connected 
with  the  First  Book  than  with  any  other.  It  is  useful  for  explaining  the 
nature  of  Hadley's  sextant;  and,  though  involved  in  the  explanations  usual- 
ly given  of  that  instrument,  it  has  not,  I  believe,  been  hitherto  considered  as 
a  distinct  Geometrical  Proposition,  though  very  well  entitled  to  be  so  on  ac- 
count of  its  simplicity  and  elegance,  as  well  as  its  utility. 

THEOREM. 

If  an  exterior  angle  of  a  triangle  be  bisected,  and  also  one  of  the  interior 
and  opposite,  the  angle  contained  by  the  bisecting  lines  is  equal  to  half  the 
other  interior  and  opposite  angle  of  the  triangle. 

Let  the  exterior  angle  ACD  of  the  triangle  ABC  be  bisected  by  the 
straight  line  CE,  and  the  interior  and  opposite  ABC  by  the  straight  line 
BE,  the  angle  BEC  is  equal  to  half  the  angle  BAG. 

The  line  CE,  BE  will  meet ;  for  since  the  angle  ACD  is  greater  than 
ABC,  the  half  of  ACD  is  greater  than  the  half  of  ABC,  that  is,  ECD 
is  greater  than  EBC  ;  add 

ECB  to  both,  and  the  two  ^E 

angles    ECD,    ECB    are  A 

greater  than  EBC,  ECB. 
But  ECD,  ECB  are  equal 
to  two  right  angles  ;  there- 
fore ECB,  EBC  are  less 
than  two  right  angles,  and 
therefore  the  lines  CE,  BE 
must  meet  on  the  same  side 
of  BC  on  which  the  trian 
gle  ABC  is.     Let  them  meet  in  E. 

Because  DCE  is  the  exterior  angle  of  the  triangle  BCE,  it  is  equal  to 
the  two  angles  CBE,  BEC,  and  therefore  twice  the  angle  DCE,  that  is,  the 
angle  DCA  is  equal  to  twice  the  angles  CBE  and  BEC.  But  twice  the 
angle  CBE  is  equal  to  the  angle  ABC,  therefore  the  angle  DCA  is  equal 
to  the  angle  ABC,  together  with  twice  the  angle  BEC  ;  and  the  same  an- 


NOTES.  299 

gle  DCA  being  the  exterior  angle  of  the  triangle  ABC,  is  equal  to  the  two 
angles  ABC,  CAB,  wherefore  the  two  angles  ABC,  CAB  are  equal  to 
ABC  and  twice  BEC.  Therefore,  taking  away  ABC  from  both,  there 
remains  the  angle  CAB  equal  to  twice  the  angle  BEC,  or  BEC  equal  to 
the  half  of  BAC, 


BOOK  II. 


The  Demonstrations  of  this  Book  are  no  otherwise  changed  than  by  in- 
troducing into  them  some  characters  similar  to  those  of  Algebra,  which  is 
always  of  great  use  where  the  reasoning  turns  on  the  addition  or  subtrac- 
tion of  rectangles.  To  Euclid's  demonstrations,  others  are  sometimes  add- 
ed, as  Scholiums,  in  which  the  properties  of  the  sections  of  lines  are  easily 
demonstrated  by  Algebraical  formulas. 


BOOK  III. 


DEFINITIONS. 

The  definition  which  Euclid  makes  the  first  of  this  Book  is  that  of  equal 
circles,  which  he  defines  to  be  "  those  of  which  the  diameters  are  equal." 
This  is  rejected  from  among  the  definitions,  as  being  a  Theorem,  the  truth 
of  which  is  proved  by  supposing  the  circles  applied  to  one  another,  so  that 
their  centres  may  coincide,  for  the  whole  of  the  one  must  then  coincide  with 
the  whole  of  the  other.  The  converse,  viz.  That  circles  which  are  equal 
have  equal  diameters,  is  proved  in  the  same  way. 

The  definition  of  the  angle  of  a  segment  is  also  omitted,  because  it  does 
not  relate  to  a  rectilineal  angle,  but  to  one  understood  to  be  contained  be- 
tween a  straight  line  and  a  portion  of  the  circumference  of  a  circle.  In  like 
manner,  no  notice  is  taken  in  the  16th  proposition  of  the  angle  comprehend- 
ed between  the  semicircle  and  the  diameter,  which  is  said  by  Euclid  to  be 
greater  than  an  acute  rectilineal  angle.  The  reason  for  these  omissions  has 
already  been  assigned  in  the  notes  on  the  fifth  definition  of  the  first  Book 

PROP.  XX. 

It  has  been  remarked  of  this  demonstration,  that  it  takes  for  granted,  that 
if  two  magnitudes  be  double  of  two  others,  each  of  each,  the  sum  or  difier- 
ence  of  the  first  two  is  double  of  the  sum  or  difference  of  the  other  two, 
which  are  two  cases  of  the  1st  and  5th  of  the  5th  Book.     The  justness  of 


# 


300  NOTES. 

this  remark  cannot  be  denierl ;  and  though  the  cases  of  the  Propositions  here 
referred  to  are  the  simplest  of  any,  yet  the  truth  of  them  ought  not  in  strict- 
ness to  be  assumed  without  proof.  The  proof  is  easily  given.  Let  A  and 
B,  C  and  D  be  four  magnitudes,  such  that  A=2C,  and  B=2D  ;  then  A 
+  B=2(C  +  D).  For  since  A  =  C4-C,  and  B  =  D  +  D,  adding  equals  to 
equals,  A  +  B=(C  +  D)  +  (C  +  D)=2(C  +  D).  So  also,  if  A  be  greater 
than  B,  and  therefore  C  greater  than  D,  since  A  =  C-f  C,  and  B=D+D, 
taking  equals  from  equals,  A — B=(C — D)  +  (C — D),  that  is,  A — B=2 
(C-D). 


BOOK  V. 

The  subject  of  proportion  has  been  treated  so  differently  by  those  who 
have  written  on  elementary  geometry,  and  the  method  which  Euclid  has  fol- 
lowed has  been  so  often,  and  so  inconsiderately  censured,  that  in  these  notes 
it  will  not  perhaps  be  more  necessary  to  account  for  the  changes  that  I  have 
made,  than  for  those  that  I  have  not  made.  The  changes  are  but  hw,  and 
relate  to  the  language,  not  to  the  essence  of  the  demonstrations  ;  they  will 
be  explained  after  some  of  the  definitions  have  been  particularly  considered. 

DEF.  III. 

The  definition  of  ratio  given  here  has  been  greatly  extolled  by  some  au- 
thors ;  but  whatever  value  it  may  have  in  the  eyes  of  a  metaphysician,  it 
has  but  little  in  those  of  the  geometer,  because  nothing  concerning  the  pro- 
perties of  ratios,  can  be  deduced  from  it.  Dr.  Barrow  has  very  judiciously 
remarked  concerning  it,  "  that  Euclid  had  probably  no  other  design  in  mak- 
*'  ing  this  definition,  than  to  give  a  general  summary  idea  of  ratio  to  begin- 
*'  ners,  by  premising  this  metaphysical  definition  to  the  more  accurate  defi- 
"  nitions  of  ratios  that  are  equal  to  one  another,  or  one  of  which  is  greater 
*'  or  less  than  the  other  ;  I  call  it  a  metaphysical,  for  it  is  not  properly  a  ma- 
"  thematical  definition,  since  nothing  in  mathematics  depends  on  it,  or  is  de- 
"  duced,  nor,  as  I  judge,  can  be  deduced,  from  it."  (Barrow's  Lectures, 
Lect.  3.)  Dr.  Simson  thinks  the  definition  has  been  added  by  some  unskil- 
ful editor ;  but  there  is  no  ground  for  that  supposition,  other  than  what  ari- 
ses from  the  definition  being  of  no  use.  We  may,  however,  well  enough 
imagine,  that  a  certain  idea  of  order  and  method  induced  Euclid  to  give 
some  general  definition  of  ratio  before  he  used  the  terra  in  the  definition  of 
equal  ratios. 

DEF.  IV. 

This  definition  is  a  little  altered  in  the  expression ;  Euclid  has  it,  that 
"  magnitudes  are  said  to  have  a  ratio  to  one  another,  when  the  less  can  be 
"  nmltiplied  so  as  to  exceed  the  greater." 


NOTES.  301 


DEF.  V. 


One  of  the  chief  obstacles  to  the  ready  understanding  of  the  5th  Book  of 
Euclid,  is  the  difficulty  that  most  people  find  of  reconciling  the  idea  of  pro- 
portion which  they  have  already  acquired,  with  the  account  of  it  that  is 
given  in  this  definition.  Our  first  ideas  of  proportion,  or  of  proportionality, 
are  got  by  trying  to  compare  together  the  magnitude  of  external  bodies  ; 
and  though  they  be  at  first  abundantly  vague  and  incorrect,  they  are  usually 
rendered  tolerably  precise  by  the  study  of  arithmetic  ;  from  which  we  learn 
to  call  four  numbers  proportionals,  when  they  are  such  that  the  quotient 
which  arises  from  dividing  the  first  by  the  second,  (according  to  the  com- 
mon rule  for  division),  is  the  same  with  the  quotient  that  arises  from  divid- 
ing the  third  by  the  fourth. 

Now,  as  the  operation  of  arithmetical  division  is  applicable  as  readily  to 
any  two  magnitudes  of  the  same  kind,  as  to  two  numbers,  the  notion  of  pro- 
portion thus  obtained  may  be  considered  as  perfectly  general.  For,  in  arith- 
metic, after  finding  how  often  the  divisor  is  contained  in  the  dividend,  we 
multiply  the  remainder  by  10,  or  100,  or  1000,  or  any  power,  as  it  is  called, 
of  10,  and  proceed  to  inquire  how  oft  the  divisor  is  contained  in  this  new 
dividend  ;  and,  if  there  be  any  remainder,  we  go  on  to  multiply  it  by  10, 
100,  &c.  as  before,  and  to  divide  the  product  by  the  original  divisor,  and  so 
on,  the  division  sometimes  terminating  when  no  remainder  is  left,  and  some- 
times going  on  ad  infinitum,  in  consequence  of  a  remainder  being  left  at  each 
operation.  Now,  this  process  may  easily  be  imitated  with  any  two  mag- 
nitudes A  and  B,  providing  they  be  of  the  same  kind,  or  such  that  the  one 
can  be  multiplied  so  as  to  exceed  the  other.  For,  suppose  that  B  is  the 
least  of  the  two ;  take  B  out  of  A  as  oft  as  it  can  be  found,  and  let  the  quo- 
tient be  noted,  and  also  the  remainder,  if  there  be  any  ;  multiply  this  remain- 
der by  10,  or  100,  &;c.  so  as  to  exceed  B,  and  let  B  be  taken  out  of  the  quan- 
tity produced  by  this  multiplication  as  oft  as  it  can  be  found  ;  let  the  quotient 
be  noted,  and  also  the  remainder,  if  there  be  any.  Proceed  with  this  remain- 
der as  before,  and  so  on  continually ;  and  it  is  evident,  that  we  have  an  opera- 
tion that  is  applicable  to  all  magnitudes  whatsoever,  and  that  maybe  perform- 
ed with  respect  to  any  two  lines,  any  two  plane  figures,  or  any  two  solids,  &c. 

Now,  when  we  have  two  magnitudes  and  two  others,  and  find  that  the 
first  divided  by  the  second,  according  to  this  method,  gives  the  very  same 
series  of  quotients  that  the  third  does  when  divided  by  the  fourth,  we  say  of 
these  magnitudes,  as  we  did  of  the  numbers  above  described,  that  the  first 
is  to  the  second  as  the  third  to  the  fourth.  There  are  only  two  more  cir- 
cumstances necessary  to  be  considered,  in  order  to  bring  us  precisely  to 
Euclid's  definition. 

First,  It  is  known  from  arithmetic,  that  the  multiplication  of  the  succes- 
sive remainders  each  of  them  by  10,  is  equivalent  to  multiplying  the  quantity 
to  be  divided  by  the  product  of  all  those  tens  ;  so  that  multiplying,  for  in- 
stance, the  first  remainder  by  10,  the  second  by  10,  and  the  third  by  10,  is 
the  same  thing,  with  respect  to  the  quotient,  as  if  the  quantity  to  be  divided 
had  been  at  first  multiplied  by  1000  ;  and  therefore,  our  standard  of  the  pro- 
portionality of  numbers  may  be  expressed  thus  :  If  the  first  multiplied  any 
number  of  times  by  1 0,  and  then  divided  by  the  second,  gives  the  same  quo- 


3(«  NOTES. 

tient  as  when  the  third  is  muliplied  as  often  by  10,  and  then  divided  by  the 
fourth,  the  four  magnitudes  are  proportionals. 

A^ain,  it  is  evident,  that  there  is  no  necessity  in  these  mukiplications  for 
confining  ourselves  to  10,  or  the  powers  of  10,  and  that  we  do  so,  in  arith- 
metic, only  for  the  conveniency  of  the  decimal  notation  ;  we  may  therefore 
use  any  multipliers  whatsoever,  providing  we  use  the  same  in  both  cases. 
Hence,  we  have  this  definition  of  proportionals,  When  there  are  four  mag- 
nitudes, and  any  multiple  whatsoever  of  the  first,  when  divided  by  the 
second,  gives  the  same  quotient  with  the  like  multiple  of  the  third,  when 
divided  by  the  fourth,  the  four  magnitudes  are  proportionals,  or  the  first 
has  the  same  ratio  to  the  second  that  the  third  has  to  the  fourth. 

We  are  now  arrived  very  nearly  at  Euclid's  definition  ;  for,  let  A,  B,  C, 
D  be  four  proportionals,  according  to  the  definition  just  given,  and  rn  any 
number  ;  and  let  the  multiple  of  A  by  m,  that  is  mA,  be  divided  by  B  ;  and 
first,  let  the  quotient  be  the  number  n  exactly,  then  also,  when  wC  is  divided 
by  D,  the  quotient  will  be  n  exactly.  But  when  /nA  divided  by  B  gives  n 
for  the  quotient,  wA=nB  by  the  nature  of  division,  so  that  when  ?»A=nB, 
»iC=nD,  which  is  one  of  the  conditions  of  Euclid's  definition. 

Again,  when  wiA  is  divided  by  B,  let  the  division  not  be  exactly  perform- 
ed, but  let  w  be  a  whole  number  less  than  the  exact  quotient,  then  nB  ^ 
>nA,  or  mA/wB  ;  and,  for  the  same  reason,  wiC/nD,  which  is  another  of 
the  conditions  of  Euclid's  definition. 

Lastly,  when  mk.  is  divided  by  B,  let  n  be  a  whole  number  greater  than 
the  exact  quotient,  then  wA/nB,  and  because  n  is  also  greater  than  the 
quotient  of  mC  divided  by  D,  (which  is  the  same  with  the  other  quotient), 
therefore  mQ> /_nY). 

Therefore,  uniting  all  these  three  conditions,  vre  call  A,  B,  C,  D,  propor- 
tionals, when  they  are  such,  that  if  mkyri^,  wiC /nD  ;  if  wiA=nB, mC=s 
nD  ;  and  if  wA^^nB,  wzC^^riD,  m  and  n  being  any  numbers  whatsoever. 
Now,  this  is  exactly  the  criterion  of  proportionality  established  by  Euclid  in 
the  5th  definition,  and  is  derived  here  by  generalizing  the  common  and  most 
familiar  idea  of  proportion. 

It  appears  from  this,  that  the  condition  of  mk  containing  B,  whether 
with  or  without  a  remainder,  as  often  aswC  contains  D,  with  or  without  a 
remainder,  and  of  this  being  the  case  whatever  value  be  assigned  to  the 
number  m,  includes  in  it  all  the  three  conditions  that  are  mentioned  in  Eu- 
clid's definition  ;  and  hence,  that  definition  may  be  expressed  a  little  more 
simply  by  saying,  that/oz/r  magnitudes  are  proportionals,  when  any  multiple  of 
the  first  contains  the  second,  {with  or  without  remainder,)  as  oft  as  the  same  mul- 
tiple of  the  third  contains  the  fourth.  But,  though  this  definition  is  certainly, 
in  the  expression,  more  simple  than  Euclid's,  it  is  not,  as  will  be  found  on 
trial,  so  easily  applied  to  the  purpose  of  demonstration.  The  three  conditions 
which  Euclid  brings  together  in  his  definition,  though  they  somewhat  em- 
barrass the  expression  of  it,  have  the  advantage  of  rendering  the  demon- 
strations more  simple  than  they  would  otherwise  be,  by  avoiding  all  discus- 
sion about  the  magnitude  of  the  remainder  left,  after  B  is  taken  out  of  twA  as 
oft  as  it  can  be  found.  All  the  attempts,  indeed,  that  have  been  made  to  de- 
monstrate the  properties  of  proportionals  rig-orously,  by  means  of  other  defini- 
tions than  Euclid's,  only  serve  to  evince  the  excellence  of  the  method  follow- 
ed by  the  Greek  Geometer,  and  his  singular  address  in  the  application  of  it 


NOTES.  303 

The  great  objection  to  the  other  methods  is,  that  if  they  are  meant  to  be 
rigorous,  they  require  two  demonstrations  to  every  proposition,  one  when 
the  division  of  mk  into  parts  equal  to  B  can  be  exactly  performed,  the  other 
when  it  cannot  be  exactly  performed  whatever  value  be  assigned  to  wi,  or 
when  A  and  B  are  what  is  called  incommensurable  ;  and  this  last  case  will 
generally  be  found  to  require  an  indirect  demonstration,  or  a  reductio  ad  ah- 
surdum. 

M.  D'Alembert,  speaking  of  the  doctrine  of  proportion,  in  a  discourse 
that  contains  many  excellent  observations,  but  in  which  he  has  overlooked 
Euclid's  manner  of  treating  this  subject  entirely,  has  the  following  remark : 
"  On  ne  pent  demontrer  que  de  cette  maniere,  (la  reduction  a  absurde,)  la 
"  plupart  des  propositions  qui  regardent  les  incommensurables.  L'idee  de 
"  I'infini  entre  au  moins  implicitemens  dans  la  notion  de  ces  sortes  de  quan- 
"  tites  ;  et  comme  nous  n'avons  qu'une  idee  negative  de  I'infini,  on  ne  peut 
*'  demontrer  directement,  et  a  priori^  tout  ce  qui  concerne  I'infini  mathema- 
"  tique."     (Encyclopedie,  mot  Geometrie.) 

This  remark  sets  in  a  strong  and  just  light  the  difficulty  of  demonstrating 
the  propositions  that  regard  the  proportion  of  incommensurable  magnitudes, 
without  having  recourse  to  the  reductio  ad  ahsurdum  :  but  it  is  surprising, 
that  M.  D'Alembert,  a  geometer  no  les?  learned  than  profound,  should 
have  neglected  to  make  mention  of  Euclid's  method,  the  only  one  in  which 
the  difficulty  he  states  is  completely  overcome.  It  is  overcome  by  the  in- 
troduction of  the  idea  of  indefinitude,  (if  I  may  be  permitted  to  use  the  word), 
instead  of  the  idea  of  infinity ;  for  m  and  n,  the  multipliers  employed,  are 
supposed  to  be  indefinite,  or  to  admit  of  all  possible  values,  and  it  is  by  the 
skilful  use  of  this  condition  that  the  necessity  of  indirect  demonstrations  is 
avoided.  In  the  whole  of  geometry,  I  know  not  that  any  happier  invention 
is  to  be  found ;  and  it  is  worth  remarking,  that  Euclid  appears  in  another 
of  his  works  to  have  availed  himself  of  the  idea  of  indefinitude  with  the 
same  success,  viz.  in  his  books  of  Porisms,  which  have  been  restored  by 
Dr.  Simson,and  in  which  the  whole  analysis  turned  on  that  idea,  as  I  have 
shown  at  length  in  the  Third  Volume  of  the  Transactions  of  the  Royal  So- 
ciety of  Edinburgh.  The  investigations  of  these  propositions  were  founded 
entirely  on  the  principle  of  certain  magnitudes  admitting  of  innumerable 
values  ;  and  the  methods  of  reasoning  concerning  them  seem  to  have  been 
extremely  similar  to  those  employed  in  the  fifth  of  the  Elements.  It  is 
curious  to  remark  this  analogy  between  the  different  works  of  the  same 
author ;  and  to  consider,  that  the  skill,  in  the  conduct  of  this  very  refined 
and  ingenious  artifice,  acquired  in  treating  the  properties  of  proportionals, 
may  have  enabled  Euclid  to  succeed  so  well  in  treating  the  still  more  dif- 
ficult subject  of  Porisms.  « 

Viewing  in  this  light  Euclid's  manner  of  treating  proportion,  I  had  no 
desire  to  change  any  thing  in  the  principle  of  his  demonstrations.  I  have 
only  sought  to  improve  the  language  of  them,  by  introducing  a  concise 
mode  of  expression,  of  the  same  nature  with  that  which  we  use  in  arith- 
metic, and  in  algebra.  Ordinary  language  conveys  the  ideas  of  the  diffe- 
rent operations  supposed  to  be  perfoi-med  in  these  demonstrations  so  slowly, 
and  breaks  them  down  'into  so  many  parts,  that  they  make  not  a  sufficient 
impression  on  the  understanding.  This  indeed  will  generally  happen  when 
the  things  treated  of  are  not  represented  to  the  senses  by  Diagrams,  as 


304.  NOTES. 

they  cannot  be  when  we  reason  concerning  magnitude  in  general,  as  in  this 
part  of  the  Elements.  Here  we  ought  certainly  to  adopt  the  language  of 
arithmetic  or  algebra,  which  by  its  shortness,  and  the  rapidity  with  which 
it  places  objects  before  us,  makes  up  in  the  best  manner  possible  for  being 
merely  a  conventional  language,  and  using  symbols  that  have  no  resem- 
blance to  the  things  expressed  by  them.  Such  a  language,  therefore,  I 
have  endeavoured  to  introduce  here ;  and  I  am  convinced,  that  if  it  shall 
be  found  an  improvement,  it  is  the  only  one  of  which  the  fifth  of  Euclid  will 
admit.  In  other  respects  I  have  followed  Dr.  Simson's  edition  to  the  accu- 
racy of  which  it  would  be  difhcult  to  make  any  addition. 

In  one  thing  I  must  observe,  that  the  doctrine  of  proportion,  as  laid  down 
here,  is  meant  to  be  more  general  than  in  Euclid's  Elements.  It  is  intended 
to  include  the  properties  of  proportional  numbers  as  well  as  of  all  magni- 
tudes. Euclid  has  not  this  design,  for  he  has  given  a  definition  of  propor- 
tional numbers  in  the  seventh  Book,  very  different  from  that  of  proportional 
magnitudes  in  the  fifth;  and  it  is  not  easy  to  justify  the  logic  of  this  man- 
ner of  proceeding ;  for  we  can  never  speak  of  two  numbers  and  two  magni- 
tudes both  having  the  same  ratios,  unless  the  word  ratio  have  in  both  cases 
the  same  signification.  All  the  propositions  about  proportionals  here 
given  are  therefore  understood  to  be  applicable  to  numbers  ;  and  accord- 
ingly, in  the  eighth  Book,  the  proposition  that  proves  equiangular  parallelo- 
grams to  be  in  a  ratio  compounded  of  the  ratios  of  the  numbers  proportional 
to  their  sides,  is  demonstrated  by  help  of  the  propositions  of  the  fifth  Book. 

On  account  of  this,  the  word  quantity,  rather  than  magnitude,  ought  in  strict- 
ness to  have  been  used  in  the  enunciation  of  these  propositions,  because  wo 
employ  the  word  Quantity  to  denote  not  only  things  extended,  to  which 
alone  we  give  the  name  of  Magnitude,  but  also  numbers.  It  will  be  suffi- 
cient, however,  to  remark,  that  all  the  propositions  respecting  the  ratios  of 
magnitudes  relate  equally  to  all  things  of  which  multiples  can  be  taken,  that 
is,  to  all  that  is  usually  expressed  by  the  word  Quantity  in  its  most  extend- 
ed signification,  taking  care  always  to  observe,  that  ratio  takes  place  only 
among  like  quantities,  (See  Def.  4.) 

DEF.  X. 

The  definition  of  compound  ratio  was  first  given  accurately  by  Dr.  Simson ; 
for,  though  Euclid  used  the  term,  he  did  so  without  defining  it.  I  have 
placed  this  definition  before  those  of  duplicate  and  triplicate  ratio,  as  it  is  in 
fact  more  general,  and  as  the  relation  of  all  the  three  definitions  is  best  seen 
when  they  are  ranged  in  this  order.  It  is  then  plain,  that  two  equal  ratios 
compound  a  ratio  duplicate  of  either  of  them ;  three  equal  ratios,  a  ratio 
triplicate  of  either  of  them,  &c. 

It  was  justly  observed  by  Dr.  Simson,  that  the  expression,  compound  ratio^ 
is  introduced  merely  to  prevent  circumlocution,  and  for  the  sake  principally 
of  enunciating  those  propositions  with  conciseness  that  are  demonstrated  by 
reasoning  ex  (pquo,  that  is,  by  reasoning  from  the  22d  or  23d  of  this  Book. 
This  will  be  evident  to  any  one  who  considers  carefully  the  Prop.  F.  of  this, 
or  the  23d  of  the  6th  Book. 

An  objection  which  naturally  occurs  to  the  use  of  the  term  compound  ratio, 
arises  from  its  not  being  evident  how  the  ratios  described  in  the  definition 


NOTES.  305 

determine  in  any  way  the  ratio  which  they  are  said  to  compounci,  since  the 
magnitudes  compounding  them  are  assumed  at  pleasure.  It  may  be  of  use 
for  removing  this  difficulty,  to  state  the  matter  as  follows  :  if  there  be  any 
number  of  ratios  (among  magnitudes  of  the  same  kind)  such  that  the  con- 
sequent of  any  of  them  is  the  antecedent  of  that  which  immediately  fol- 
lows, the  first  of  the  antecedents  has  to  the  last  of  the  consequents  a  ratio 
which  evidently  depends  on  the  intermediate  ratios,  because  if  they  are  de- 
termined, it  is  determined  also  ;  and  this  dependence  of  one  ratio  on  all  the 
other  ratios,  is  expressed  by  saying  that  it  is  compounded  of  them.     Thus, 

if  _.j        _-_j  -p^jbe  any  series  of  ratios,  such  as  described  above,  the  ratio 

j3    C    D    JE 
A  A    R 

— ,  or  of  A  to  E,  is  said  to  be  compounded  of  the  ratios  — ,  -^,  &c.    The  ratio 

A  A     TJ 

-=i-j  is  evidently  determined  by  the  ratios  — ,  — ,  &c.  because  if  each  of  the 
E  Jd     O 

latter  is  fixed  and  invariable,  the  former  cannot  change.     The  exact  nature 

of  this  dependence,  and  how  the  one  thing  is  determined  by  the  other,  it  is 

not  the  business  of  the  definition  to  explain,  but  merely  to  give  a  name  to 

a  relation  which  it  may  be  of  importance  to  consider  more  attentively. 


BOOK  VI. 

DEFINITION  II. 


This  definition  is  changed  from  that  of  reciprocal  figures  ^  which  was  of  no 
use,  to  one  that  corresponds  to  the  language  used  in  the  14th  and  15th 
propositions,  and  in  other  parts  of  geometry. 

PROP.  A,  B,  C,  &c. 

Nine  propositions  are  added  to  this  Book  on  account  of  their  utility  and 
their  connection  with  this  part  of  the  Elements.  The  first  four  of  them  are 
in  Dr.  Simson's  edition,  and  among  these  Prop.  A  is  given  immediately 
after  the  third,  being,  in  fact,  a  second  case  of  that  proposition,  and  capable 
of  being  included  with  it,  in  one  enunciation.  Prop.  D  is  remarkable  for 
being  a  theorem  of  Ptolemy  the  Astronomer,  in  his  MsyaXr]  2vvialig,  and  the 
foundation  of  the  construction  of  his  trigonometrical  tables.  Prop.  E  is  the 
simplest  case  of  the  former  ;  it  is  also  useful  in  trigonometry,  and,  under 
another  form,  was  the  97th,  or,  in  some  editions,  the  94th  of  Euclid's  Data. 
The  propositions  F  and  G  are  very  useful  properties  of  the  circle,  and  are 
taken  from  the  Loci  Plani  of  ApoUonius.  Prop.  H  is  a  very  remarkable  pro- 
perty of  the  triangle  ;  and  K  is  a  proposition  which,  though  it  has  been 
hitherto  considered  as  belonging  particularly  to  trigonometry,  is  so  often  of 
use  in  other  parts  of  the  mathematics,  that  it  may  be  properly  ranked  among 
elementary  theorems  of  Geometry. 

39 


SUPPLEMENT. 


BOOK  I. 

PROP.  V.  and  VI,  &c. 


The  demonstrations  of  the  5th  and  6th  propositions  require  the  method 
of  exhaustions,  that  is  to  say,  they  prove  a  certain  property  to  belong  to  the 
circle,  because  it  belongs  to  the  rectilineal  figures  inscribed  in  it,  or  described 
about  it  according  to  a  certain  law,  in  the  case  when  those  figures  ap- 
proach to  the  circles  so  nearly  as  not  to  fall  short  of  it  or  to  exceed  it,  by 
any  assignable  difference.  This  principle  is  general,  and  is  the  only  one 
by  which  we  can  possibly  compare  curvilineal  with  rectilineal  spaces^  or  the 
length  of  curve  lines  with  the  length  of  straight  lines,  whether  we  follow 
the  methods  of  the  ancient  or  of  the  modern  geometers.  It  is  therefore  a 
great  injustice  to  the  latter  methods  to  represent  them  as  standing  on  a  foun- 
dation less  secure  than  the  former ;  thfcy  stand  in  reality  on  the  same,  and 
the  only  difference  is,  that  the  application  of  the  principle,  common  to  thena 
both,  is  more  general  and  expeditious  in  the  one  case  than  in  the  other. 
This  identity  of  principle,  and  affinity  of  the  methods  used  in  the  elementary 
and<he  higher  mathematics,  it  seems  the  most  necessary  to  observe,  that 
some  learned  mathematicians  have  appeared  not  to  be  sufficiently  aware  of 
it,  and  have  even  endeavoured  to  demonstrate  the  contrary.  An  instance 
of  this  is  to  be  met  with  in  the  preface  of  the  valuable  edition  of  the  works 
of  Archimedes,  lately  printed  at  Oxford.  In  that  preface,  Torelli,  the  learn- 
ed commentator,  whose  labours  have  done  so  much  to  elucidate  the  writ- 
ings of  the  Greek  Geometer,  but  who  is  so  unwilling  to  acknowledge  the 
merit  of  the  modern  analysis,  undertakes  to  prove,  that  it  is  impossible,  from 
the  relation  which  the  rectilineal  figures  inscribed  in,  and  circumscribed 
about,  a  given  curve  have  to  one  another,  to  conclude  any  thing  concerning 
the  properties  of  the  curvilineal  space  itself,  except  in  certain  circumstances 
which  he  has  not  precisely  described.  "With  this  view  he  attempts  to  show, 
that  if  we  are  to  reason  from  the  relation  which  certain  rectilineal  figures 
belonging  to  the  circle  have  to  one  another,  notwithstanding  that  those 
figures  may  approach  so  near  to  the  circular  spaces  within  which  they  are 
inscribed,  as  not  to  differ  from  them  by  any  assignable  magnitude,  we  shall 
be  led  into  error,  and  shall  seem  to  prove,  that  the  circle  is  to  the  square  of 
its  diameter  exactly  as  3  to  4.  Now,  as  this  is  a  conclusion  which  the  dis- 
coveries of  Archimedes  himself  prove  so  clearly  to  be  false,  Torelli  argues, 
that  the  principle  from  which  it  is  deduced  must  be  false  also  ;  and  in  this 
he  would  no  doubt  be  right,  if  his  former  conclusion  had  been  fairly  drawn. 
But  the  truth  is,  that  a  very  gross  paralogism  is  to  be  found  in  that  part  of 


NOTES.  SUPPL.  BOOK  I.  307 

his  reasoning,  where  he  makes  a  transition  from  the  ratios  of  the  small  rect- 
angles, inscribed  in  the  circular  spaces,  to  the  ratios  of  the  sums  of  those 
rectangles,  or  of  the  whole  rectilineal  figures.  In  doing  this,  he  takes  for 
granted  a  proposition,  which,  it  is  wonderful,  that  one  who  had  studied 
geometry  in  the  school  of  Archimedes,  should  for  a  moment  have  suppos- 
ed to  be  true.  The  proposition  is  this  :  If  A,  B,  C,  D,  E,  F,  be  any  num- 
ber of  magnitudes,  and  a,  b,  c,  d,  e,f,  as  many  others  ;  and  if 
A  :  B  : ;  (z  :  ft, 
C:D  ::c:  d, 

E  :  F  :  :  «  :  /,  then  the  sum  of  A,  C  and  E  will  be  to  the  sum  of  B,  D  and 
F,  as  the  sum  of  a,  c  and  e,  to  the  sum  of  b,  d  and/,  or  A-f-C+E  :  B-|-D 
-f-F  :  :  a-fc+e  :  b-\-d-{-f.  No  w,  this  proposition,  which  Torelli  supposes 
to  be  perfectly  general,  is  not  true,  except  in  two  cases,  viz.  either  first, 
when  A  :  C  :  :  a  :  c,  and 

A  :  E  :  :  a  :  e  ;  and  consequently, 
B  :  D  :  :  5  :  rf,  and 

B  :  F  :  :  b  :  f;  or,  secondly,  when  all  the  ratios  of  A  to  B,  C  to  D,  E 
to  F,  &c.  are  equal  to  one  another.  To  demonstrate  this,  let  us  suppose 
that  there  are  four  magnitudes,  and  four  others, 

thus  A  :  B  :  :  a  :  bj  and 

C  :  D  :  :  c  :  dj  then  we  cannot  have 
A+C  :  B  +  D  : :  a-\-c  :  5-f(?,  unless  either  A  :  C  ; :  a  :  c,  andB  :D  : :  b  : 
d ;  or  A  :  C  :  :  b  :  d,  and  consequently  a  :  b  :  :  c  :  d. 

Take  a  magnitude  K,  such  that  a  :  c  : :  A  :  K,  and  another  L,  such  that 
h  :  d  :  :  B  :  h  ;  and  suppose  it  true,  that  A  +  C  :  B  +  D  :  : 
a+c  :  b-{-d.  Then,  because  by  inversion  ;  K  :  A  :  :  c  :  a, 
and,  by  hypothesis,  A  :  B  :  :  a  :  b,  and  also  B  :  L  : :  b  :  dy 
ex  aequo,  K  :L  : :  c:  d;    and  consequently,  K  :  L  : : 
C:D. 

Again,  because  A  :  K  : :  c  :  c,  by  addition, 

A-i-K  :  K  : :  a-^-c  :  c ;  and  for  the  same  reason, 
B-j-L  :  L  :  :  b'{-d  :  d,  or,  by  inversion, 
L  :  B-\-h  :  :  d  :  b+d.     And,  since  it  has  been  shewn,  that 
K  :  h  : :  c  :  d;  therefore,  ex  aequo. 


K,A,B,L, 
c,    a,  by  d. 


A+K,K,L,B+L, 
a-\-Cy     c,  d,  b-{'d. 


A-i-K  :  B+L  :  :  a+c  :  b-\-d;  but  by  hypothesis, 

A-f-C  :  B  +  D  :  :  a+c  :  b+d,  therefore 

A+K  :  A+C  : :  B+L  :  B-f  D. 
Now,  first,  let  K  and  C  be  supposed  equal,  then  it  is  evident  that  L  and 
D  are  also  equal ;  and  therefore,  since  by  construction  a  :  c  : :  A  :  K,  we 
have  also  a  :  c  :  :  A  :  C  ;  and,  for  the  same  reason,  5  :  J  : :  B  :  D,  and 
these  analogies  from  the  first  of  the  two  conditions,  of  which  one  is  affirmed 
above  to  be  always  essential  to  the  truth  of  Torelli's  proposition. 
Next,  if  K  be  greater  than  C,  then,  since 

A-fK  :  A+C  :  :  B+L  :  B+D,  by  division, 

A-fK  :  K— C  : :  B+L  :  L—D.     But,  as  was  shewn, 

K  :  L  :  :  C  :  D,  by  conversion  and  alternation, 

K— C  :  K  : ;  L—D  :  L,  therefore,  ex  aequo. 


308  NOTES.  SUPPL.  BOOK  II. 

A+K  ;  K  :  :  B+L  :  L>  and  lastly,  by  division, 
A  :  K  : :  B  :  L,  or  A  :  B  : ;  K  :  L,  that  is, 
A  :  B  ;  :  C  :  D. 

"Wherefore,  in  this  case  the  ratio  of  A  to  B  is  equal  to  that  of  C  to  D, 
and  consequently,  the  ratio  of  a  to  h  equal  to  that  of  c  to  d.  The  same 
may  be  shewn,  if  K  is  less  than  C  ;  therefore  in  every  case  there  are  con- 
ditions necessary  to  the  truth  of  Torelli's  proposition,  which  he  does  not 
take  into  account,  and  which,  as  is  easily  shewn,  do  not  belong  to  the  mag- 
nitudes to  which  he  applies  it. 

In  consequence  of  this,  the  conclusion  which  he  meant  to  establish  re- 
specting the  circle,  falls  entirely  to  the  ground,  and  with  it  the  general  in- 
ference aimed  against  the  modern  analysis. 

It  will  not,  I  hope,  be  imagined,  that  I  have  taken  notice  of  these  cir- 
cumstances with  any  design  to  lessen  the  reputation  of  the  learned  Italian, 
who  has  in  so  many  respects  deserved  well  of  the  mathematical  sciences, 
or  to  detract  from  the  value  of  a  posthumous  work,  which  by  its  elegance 
and  correctness,  does  so  much  honour  to  the  English  editors.  But  I  would 
warn  the  student  against  that  narrow  spirit  which  seeks  to  insinuate  itself 
even  into  the  abstractions  of  geometry,  and  would  persuade  us,  that  ele- 
gance, nay,  truth  itself,  are  possessed  exclusively  by  the  ancient  methods 
of  demonstration.  The  high  tone  in  which  Torelli  censures  the  modern  ma- 
thematics is  imposing,  as  it  is  assumed  by  one  who  had  studied  the  writings 
of  Archimedes  with  uncommon  diligence.  His  errors  are  on  that  account 
the  more  dangerous,  and  require  to  be  the  more  carefully  pointed  out. 

PROP.  IX. 

This  enunciation  is  the  same  with  that  of  the  third  of  the  Dimensio  Ctr^ 
cull  of  Archimedes  ;  but  the  demonstration  is  different,  though  it  proceeds 
like  that  of  the  Greek  Geometer,  by  the  continual  bisection  of  the  6th  part 
of  the  circumference. 

The  limits  of  the  circumference  are  thus  assigned ;  and  the  method  of 
bringing  it  about,  notwithstanding  many  quantities  are  neglected  in  the  arith- 
metical operations,  that  the  errors  shall  in  one  case  be  all  on  the  side  of  de- 
fect, and  in  another  all  on  the  side  of  excess  (in  which  I  have  followed  Ar- 
chimedes,) deserves  particularly  to  be  observed,  as  affording  a  good  intro- 
duction to  the  general  methods  of  approximation. 


BOOK  IL 

DEF.  VIII.  and  PROP.  XX. 

Solid  angles,  which  are  defined  here  in  the  same  manner  as  in  Euclid, 
are  magnitudes  of  a  very  peculiar  kind,  and  are  particularly  to  be  remarked 
for  not  admitting  of  that  accurate  comparison,  one  with  another,  which  is 


NOTES.  SUPPL.  BOOK  II.  309 

common  in  the  other  subjects  of  geometrical  investigation.  It  cannot,  for 
example,  be  said  of  one  solid  angle,  that  it  is  the  half,  or  the  double  of  an- 
other solid  angle  ;  nor  did  any  geometer  ever  think  of  proposing  the  pro- 
blem of  bisecting  a  given  solid  angle.  In  a  word,  no  multiple  or  sub-mul- 
tiple of  such  an  angle  can  be  taken,  and  we  have  no  way  of  expounding, 
even  to  the  simplest  cases,  the  ratio  which  one  of  them  bears  to  another. 

In  this  respect,  therefore,  a  solid  angle  differs  from  every  other  magni- 
tude that  is  the  subject  of  mathematical  reasoning,  all  of  which  have  this 
common  property,  that  multiples  and  sub-multiples  of  them  may  be  found. 
It  is  not  our  business  here  to  inquire  into  the  reason  of  this  anomaly,  but  it 
is  plain,  that  on  account  of  it,  our  knowledge  of  the  nature  and  the  proper- 
ties of  such  angles  can  never  be  very  far  extended,  and  that  our  reason- 
ings concerning  them  must  be  chiefly  confined  to  the  relations  of  the  plane 
angles,  by  which  they  are  contained.  One  of  the  most  remarkable  of  those 
relations  is  that  which  is  demonstrated  in  the  21st  of  this  Book,  and  which 
is,  that  all  the  plane  angles  which  contain  any  solid  angle  must  together 
be  less  than  four  right  angles.  This  proposition  is  the  21st  of  the  11th  of 
Euclid. 

This  proposition,  however,  is  subject  to  a  restriction  in  certain  cases, 
which,  I  believe,  was  first  observed  by  M.  le  Sage  of  Geneva,  in  a  com- 
munication to  the  Academy  of  Sciences  of  Paris  in  1756.  When  the  sec- 
tion of  the  pyramid  formed  by  the  planes  that  contain  the  solid  angle  is  a 
figure  that  has  none  of  its  angles  exterior,  such  as  a  triangle,  a  parallelo- 
gram, &;c.  the  truth  of  the  proposition  just  enunciated  cannot  be  question- 
ed. But,  when  the  aforesaid  section  is  a  figure  like  that  which  is  annexed, 
viz.  ABCD,  having  some  angles  such 
as  BDC,  exterior,  or,  as  they  are  some- 
times called,  re-entering  angles,  the 
proposition  is  not  necessarily  true  ; 
and  it  is  plain,  that  in  such  cases  the 
demonstration  which  we  have  given, 
and  which  is  the  same  with  Euclid's, 
will  no  longer  apply.  Indeed,  it  were 
easy  to  show,  that  on  bases  of  this 
kind,  by  multiplying  the  number  of 
sides,  solid  angles  maybe  formed,  such 
that  the  plane  angles  which  contain  them  shall  exceed  four  right  angles  by 
any  quantity  assigned.  An  illustration  of  this  from  the  properties  of  the 
sphere  is  perhaps  the  simplest  of  all  others.  Suppose  that  on  the  surface 
of  a  hemisphere  there  is  described  a  figure  bounded  by  any  number  of  arcs 
of  great  circles  making  angles  with  one  another,  on  opposite  sides  alter- 
nately, the  plane  angles  at  the  centre  of  the  sphere  that  stand  on  these  arcs 
may  evidently  exceed  four  right  angles,  and  that  too,  by  multiplying  and 
extending  the  arcs  in  any  assigned  ratio.  Now,  these  plane  angles  con- 
tain a  solid  angle  at  the  centre  of  the  sphere,  according  to  the  definition  of 
a  solid  angle. 

We  are  to  understand  the  proposition  in  the  text,  therefore,  to  be  true 
only  of  those  solid  angles  in  which  the  inclination  of  the  plane  angles  are 
all  the  same  way,  or  all  directed  toward  the  interior  of  the  figure.  To  dis- 
tinguish tliis  class  of  solid  angles  from  that  to  which  the  proposition  does 


310  NOTES.  SUPPL.  BOOK  II. 

not  apply,  it  is  perhaps  best  to  make  use  of  this  criterion,  that  they  are  such, 
that  when  any  two  points  whatsoever  are  taken  in  the  planes  that  contain 
the  solid  angle,  the  straight  line,  joining  those  points,  falls  wholly  within 
the  solid  angle  :  or  thus,  they  are  such,  that  a  straight  line  cannot  meet  the 
planes  which  contain  them  in  more  than  two  points.  It  is  thus,  too,  that  I 
would  distinguish  a  plane  figure  that  has  none  of  its  angles  exterior,  by 
saying,  that  it  is  a  rectilineal  figure,  such  that  a  straight  line  cannot  meet 
the  boundary  of  it  in  more  than  two  points. 

We,  therefore,  distinguish  solid  angles  into  two  species :  one  in  which 
the  bounding  planes  can  be  intersected  by  a  straight  line  only  in  two 
points  ;  and  another  where  the  bounding  planes  may  be  intersected  by  a 
straight  line  in  more  than  two  points :  to  the  first  of  these  the  proposition 
in  the  text  applies,  to  the  second  it  does  not. 

Whether  Euclid  meant  entirely  to  exclude  the  consideration  of  figures 
of  the  latter  kind,  in  all  that  he  has  said  of  solids,  and  of  solid  angles,  it  is 
not  now  easy  to  determine  :  it  is  certain,  that  his  definitions  involve  no 
such  exclusion  ;  and  as  the  introduction  of  any  limitation  would  conside- 
rably embarrass  these  definitions,  and  render  them  difficult  to  be  understood 
by  a  beginner,  I  have  left  it  out,  reserving  to  this  place  a  fuller  explanation 
of  the  difficulty.  I  cannot  conclude  this  note  without  remarking,  with  the 
historian  of  the  Academy,  that  it  is  extremely  singular,  that  not  one  of  all 
those  who  had  read  or  explained  Euclid  before  M.  le  Sage,  appears  to 
have  been  sensible  of  this  mistake.  (Memoires  de  VAcad.  des  Sciences f 
1756,  Hist.  p.  77.)  A  circumstance  that  renders  this  still  more  singular 
is,  that  another  mistake  of  Euclid  on  the  same  subject,  and  perhaps  of  all 
other  geometers,  escaped  M.  le  Sage  also,  and  was  first  discovered  by 
Dr.  Simson,  as  will  presently  appear. 

PROP.  IV. 

This  very  elegant  demonstration  is  from  Legendre,  and  is  much  easier 
than  that  of  Euclid. 

The  demonstration  given  here  of  the  6th  is  also  greatly  simpler  than 
that  of  Euclid.  It  has  even  an  advantage  that  does  not  belong  to  Legen- 
dre's,  that  of  requiring  no  particular  construction  or  determination  of  any 
one  of  the  lines,  but  reasoning  from  properties  common  to  every  part  of 
them.  The  simplification,  when  it  can  be  introduced,  which,  however, 
does  not  appear  to  be  always  possible,  is,  perhaps,  the  greatest  improve- 
ment that  can  be  made  on  an  elementary  demonstration.  , 

PROP.  XIX. 

The  problem  contained  in  this  proposition,  of  drawing  a  straight  line  per- 
pendicular to  two  straight  lines  not  in  the  same  plane,  is  certainly  to  be  ac- 
counted elementary,  although  not  given  in  any  book  of  elementary  geome- 
try that  I  know  of  before  that  of  Legendre.  The  solution  given  here  is 
more  simple  than  his,  or  than  any  other  that  I  have  yet  met  with  :  it  also 
leads  more  easily,  if  it  be  required,  to  a  trigonometrical  computation. 


NOTES.    SUPPL.    BOOK  III.  311 

BOOK  m. 

DEF.  II.  and  PROP.  I. 

These  relate  to  similar  and  equal  solids,  a  subject  on  which  mistakes  have 
prevailed  not  unlike  to  that  which  has  just  been  mentioned.  The  equality 
of  solids,  it  is  natural  to  expect,  must  be  proved  like  the  equality  of  plane 
figures,  by  showing  that  they  may  be  made  to  coincide,  or  to  occupy  the 
same  space.  But,  though  it  be  true  that  all  solids  which  can  be  shewn  to 
coincide  are  equal  and  similar,  yet  it  does  not  hold  conversely,  that  all  solids 
which  are  equal  and  similar  can  be  made  to  coincide.  Though  this  asser- 
tion may  appear  somewhat  paradoxical,  yet  the  proof  of  it  is  extremely 
simple. 

Let  ABC  be  an  isosceles  triangle,  of  which  the  equal  sides  are  AB  and 
AC ;  from  A  draw  AE  perpendicular  to  the  base  BC,  and  BC  will  be  bisected 
in  E.  From  E  draw  ED  perpendicular  to  the 
plane  ABC,  and  from  D,  any  point  in  it,  draw 
DA,  DB,  DC  to  the  three  angles  of  the  tri- 
angle ABC.  The  pyramid  DABC  is  divided 
into  two  pyramids  DABE,  DACE,  which, 
though  their  equality  will  not  be  disputed, 
cannot  be  so  applied  to  one  another  as  to  coin- 
cide. For,  though  the  triangles  ABE,  ACE 
are  equal,  BE  being  equal  to  CE,  EA  common 
to  both,  and  the  angles  AEB,  AEC  equal,  be- 
cause they  are  right  angles,  yet  if  these  two 
triangles  be  applied  to  one  another,  so  as  to 
coincide,  the  solid  DACE  will  nevertheless, 
as  is  evident,  fall  without  the  solid  DABE,  for  the  two  solids  will  be  on  the 
opposite  sides  of  the  plane  ABE.  In  the  same  way,  though  all  the  planes 
of  the  pyramid  DABE  may  easily  be  shewn  to  be  equal  to  those  of  the  py- 
ramid DACE,  each  to  each ;  yet  will  the  pyramids  themselves  never  coin- 
cide, though  the  equal  planes  be  applied  to  one  another,  because  they  are 
on  the  opposite  sides  of  those  planes. 

It  may  be  said,  then,  on  what  ground  do  we  conclude  the  pyramids  to 
be  equal  ?  The  answer  is,  because  their  construction  is  entirely  the  same, 
and  the  conditions  that  determine  the  magnitude  of  the  one  identical  with 
those  that  determine  the  magnitude  of  the  other.  For  the  magnitude  of 
the  pyramid  DABE  is  determined  by  the  magnitude  of  the  triangle  ABE, 
the  length  of  the  line  ED,  and  the  position  of  ED,  in  respect  of  the  plane 
ABE  ;  three  circumstances  that  are  precisely  the  same  in  the  two  pyra- 
mids, so  that  there  is  nothing  that  can  determine  one  of  them  to  be  greater 
than  another. 

This  reasoning  appears  perfectly  conclusive  and  satisfactory ;  and  it 
seems  also  very  certain,  that  there  is  no  other  principle  equally  simple,  on 
which  the  relation  of  the  solids  DABE,  DACE  to  one  another  can  be  de- 
termined. Neither  is  this  a  case  that  occurs  rarely ;  it  is  one,  that,  in  the 
comparison  of  magnitudes  having  three  dimensions,  presents  itself  conti- 


312  NOTES.     SUPPL.     BOOK  III. 

nually  ;  for,  though  two  plane  figures  that  are  equal  and  similar  can  always 
be  made  to  coincide,  yet,  with  regard  to  solids  that  are  equal  and  similar  if 
they  have  not  a  certain  similarity  in  their  position,  there  will  be  found  just 
as  many  cases  in  which  they  cannot,  as  in  which  they  can  coincide.  Even 
figures  described  on  surfaces,  if  they  are  not  plane  surfaces,  may  be  equal 
and  similar  without  the  possibility  of  coinciding.  Thus,  in  the  fio-ure  de- 
scribed on  the  surface  of  a  sphere,  called  a  spherical  triangle,  if  we  suppose 
it  to  be  isosceles,  and  a  perpendicular  to  be  drawn  from  the  vertex  on  the 
base,  it  will  not  be  doubted,  that  it  is  thus  divided  into  two  right  angled 
spherical  triangles  equal  and  similar  to  one  another,  and  which,  neverthe- 
less, cannot  be  so  laid  on  one  another  as  to  agree.  The  same  holds  in  in- 
numerable other  instances,  and  therefore  it  is  evident,  that  a  principle,  more 
general  and  fundamental  than  thtit  of  the  equality  of  coinciding  figures, 
ought  to  be  introduced  into  Geometry.  What  this  principle  is  has  also  ap- 
peared very  clearly  in  the  course  of  these  remarks ;  and  it  is  indeed  no 
other  than  the  principle  so  celebrated  in  the  philosophy  of  Leibnitz,  under 
the  name  of  the  sufficient  reason.  For  it  was  shewn,  that  the  pyra- 
mids DABE  and  DACE  are  concluded  to  be  equal,  because  each  of  them 
is  determined  to  be  of  a  certain  magnitude,  rather  than  of  any  other,  by 
conditions  that  are  the  same  in  both,  so  that  there  is  no  reason  for  the  one 
being  greater  than  the  other.  This  Axiom  may  be  rendered  general  by 
saying,  That  things  of  which  the  magnitude  is  determined  by  conditions 
that  are  exactly  the  same,  are  equal  to  one  another ;  or,  it  might  be  ex- 
pressed thus  ;  Two  magnitudes  A  and  B  are  equal,  when  there  is  no  rea- 
son that  A  should  exceed  B,  rather  than  that  B  should  exceed  A.  Either 
of  these  will  serve  as  the  fundamental  principle  for  comparing  geometrical 
magnitudes  of  every  kind  ;  they  will  apply  in  those  cases  where  the  coin- 
cidence of  magnitudes  with  one  another  has  no  place  ;  and  they  will  apply 
with  great  readiness  to  the  cases  in  which  a  coincidence  may  take  place, 
such  as  in  the  4th,  the  8th,  or  the  26th  of  the  First  Book  of  the  Ele- 
ments. 

The  only  objection  to  this  Axiom  is,  that  it  is  somewhat  of  a  metaphy- 
sical kind,  and  belongs  to  the  doctrine  of  the  sujjicient  reason,  which  is  looked 
on  with  a  suspicious  eye  by  some  philosophers.  But  this  is  no  solid  objec- 
tion ;  for  such  reasoning  may  be  applied  with  the  greatest  safety  to  those 
objects  with  the  nature  of  which  we  are  perfectly  acquainted,  and  of  which 
we  have  complete  definitions,  as  in  pure  mathematics.  In  physical  ques 
tions,  the  same  principle  cannot  be  applied  with  equal  safety,  because  in 
such  cases  we  have  seldom  a  complete  definition  of  the  thing  we  reason 
about,  or  one  that  includes  all  its  properties.  Thus,  when  Archimedes  prov- 
ed the  spherical  figure  of  the  earth,  by  reasoning  on  a  principle  of  this  sort, 
he  was  led  to  a  false  conclusion,  because  he  knew  nothing  of  the  rotation  of 
the  earth  on  its  axis,  which  places  the  particles  of  that  body,  though  at 
equal  distances  from  the  centre,  in  circumstances  very  difierent  from  one 
another.  But,  concerning  those  things  that  are  the  creatures  of  the  mind 
altogether,  like  the  objects  of  mathematical  investigation,  there  can  be  no 
danger  of  being  misled  by  the  principle  of  the  sufficient  reason,  which  at  the 
same  time  furnishes  us  with  the  only  single  Axiom,  by  help  of  which  we 
can  compare  together  geometrical  quantities,  whether  they  be  of  one,  of 
two,  or  of  three  dimensions. 


NOTES.  SUPPL.  BOOK  HI.  313 

Legendre  in  his  Elements  has  made  the  same  remark  that  na-s  been  just 
stated,  that  there  are  solids  and  other  Geometrical  Magnitudes,  which, 
though  similar  and  equal,  cannot  be  brought  to  coincide  with  one  another, 
and  he  has  distinguished  them  by  the  name  of  Symmetrical  Magnitudes.  He 
has  also  given  a  very  satisfactory  and  ingenious  demonstration  of  the  equa- 
lity of  certain  solids  of  that  sort,  though  not  so  concise  as  the  nature  of  a 
simple  and  elementary  truth  would  seem  to  require,  and  consequently  not 
sifch  as  to  render  the  axiom  proposed  above  altogether  imnecessary 

But  a  circumstance  for  which  I  cannot  very  well  account  is,  that  Legen- 
/dre,  and  after  him  Lacroix,  ascribe  to  Simson  the  first  mention  of  such  solids 
as  we  are  here  considering.  Now  I  must  be  permitted  to  say,  that  no  re- 
jnark  to  this  purpose  is  to  be  found  in  any  of  the  writings  of  Simson,  which 
have  come  to  my  knowledge.  He  has  indeed  made  an  observation  concerning 
^  ^  the  Geometry  of  Solids,  which  was  both  new  and  important,  viz.  that  solids 
may  have  the  condition  which  Euclid  thought  sufficient  to  determine  their 
quality,  and  may  nevertheless  be  unequal ;  whereas  the  observation  made 
here  is,  that  solids  may  be  equal  and  similar,  and  may  yet  want  the  condition 
of  being  able  to  coincide  with  one  another.  These  propositions  are  widely 
different ;  and  how  so  accurate  a  writer  as  Legendre  should  have  mistaken 
the  one  for.  the  other,  is  not  easy  to  be  explained.  It  must  be  observed, 
that  he  does  pot  seem  in  the  least  aware  of  the  observation  which  Simson 
has  really  made.  Perhaps  having  himself  made  the  remark  we  now  speak 
of,  and  on  looking  slightly  into  Simson,  having  found  a  limitation  of  the 
usual  description  of  equal  solids,  he  had  without  much  inquiry,  set  it  down 
as  the  same  with  his  own  notion ;  and  so,  with  a  great  deal  of  candour, 
and  some  precipitation,  he  has  ascribed  to  Simson  a  discovery  which  really 
belonged  to  himself.  This  at  least  seems  to  b^  the  most  probable  solution 
of  the  difficulty. 

I  have  entered  into  a  fuller  discussion  of  Legendre's  mistake  than  I 
should  otherwise  have  done,  from  having  said,  in  the  first  edition  of  these 
elements,  in  1795,  that  I  believed  the  non-coincidence  of  similar  and  equal 
solids  in  certain  circumstances,  was  then  made  for  the  first  time.  This  it 
is  evident  would  have  been  a  pretension  as  ridiculous  as  ill-founded,  if  the 
same  observation  had  been  made  in  a  book  like  Simson's,  which  in  this 
country  was  in  every  body's  hands,  and  which  I  had  myself  professedly 
studied  with  attention.  As  I  have  not  seen  any  edition  of  Legendre's  Ele- 
ments earlier  than  that  published  in  1802,  I  am  ignorant  whether  he  or  I 
was  the  first  in  making  the  remark  here  referred  to.  That  circumstance 
is,  however,  immaterial ;  for  I  am  not  interested  about  the  originality  of  the 
remark,  though  very  much  interested  to  show  that  I  had  no  intenton  of  ap- 
propriating to  myself  a  discovery  made  by  another. 

Another  observation  on  the  subject  of  those  solids,  which,  with  Legendre, 
we  shall  call  Symmetrical,  has  occurred  to  me,  which  I  did  not  at  first 
think  of,  viz.  that  Euclid  himself  certainly  had  these  solids  in  view  when  he 
formed  his  definition  (as  he  very  improperly  calls  it)  of  equal  and  similar  solids. 
He  says  that  those  solids  are  equal  and  similar,  which  are  contained  under 
he  same  number  of  equal  and  similar  planes.  But  this  is  not  true,  as  Dr. 
Simson  has  shewn  in  a  passage  just  about  to  be  quoted,  because  two  solids 
may  easily  be  assigned,  bounded  by  the  same  numbe;  of  equal  and  similar 
—  planes,  which  are  obviously  unequal,  the  one  being  contained  within  the 

40 


314  NOTES.     SUPPL.    BOOK  III. 

other.  Simson  observes,  that  Euclid  needed  only  to  have  added,  that  the 
equal  and  similar  planes  must  be  similarly  situated,  to  have  made  his  des- 
cription exact.  Now,  it  is  true,  that  this  addition  would  have  made  it  exact 
in  one  respect,  but  would  have  rendered  it  imperfect  in  another ;  for  though 
all  the  solids  having  the  conditions  here  enumerated,  are  equal  and  similar, 
many  others  are  equal  and  similar  which  have  not  those  conditions,  that  is, 
though  bounded  by  the  same  equal  number  of  similar  planes,  those  planes 
are  not  similarly  situated.  The  symmetrical  solids  have  not  their  equal 
and  similar  planes  similarly  situated,  but  in  an  order  and  position  directly  con- 
trary. Euclid,  it  is  probable,  was  aware  of  this,  and  by  seeking  to  render 
the  description  of  equal  and  similar  solids  so  general,  as  to  comprehend  so- 
lids of  both  kinds,  has  stript  it  of  an  essential  condition,  so  that  solids  ob- 
viously unequal  are  included  in  it,  and  has  also  been  led  into  a  very  illogical 
proceeding,  that  of  defining  the  equality  of  solids,  instead  of  proving  it,  as  if 
he  had  been  at  liberty  to  fix  a  new  idea  to  the  word  equal  every  time  that 
he  applied  it  to  a  new  kind  of  magnitude.  The  nature  of  the  difficulty  he 
had  to  contend  with,  will  perhaps  be  the  more  readily  admitted  as  an  apo- 
logy for  this  error,  when  it  is  considered  that  Simson,  who  had  studied  the 
matter  so  carefully,  as  to  set  Euclid  right  in  one  particular,  was  himself 
wrong  in  another,  and  has  treated  of  equal  and  similar  solids,  so  as  to  ex- 
clude the  symmetrical  altogether,  to  which  indeed  he  seems  not  to  have  at 
all  adverted. 

I  must,  therefore,  again  repeat,  that  I  do  not  think  that  this  matter  can 
be  treated  in  a  way  quite  simple  and  elementary,  and  at  the  same  time 
general,  without  introducing  the  principle  of  the  sufficient  reason  as  stated 
above.  It  may  then  be  demonstrated,  that  similar  and  equal  solids  axe 
those  contained  by  the  same  number  of  equal  and  similar  planes,  either  with 
similar  or  contrary  situations.  If  the  word  contrary  is  properly  understood, 
this  description  seems  to  be  quite  general. 

Simson's  remark,  that  solids  may  be  unequal,  though  contained  by  the 
same  number  of  equal  and  similar  planes,  extends  also  to  solid  angles 
which  may  be  unequal,  though  contained  by  the  same  number  of  equal 
plane  angles.  These  remarks  he  published  in  the  first  edition  of  his  Eu- 
clid in  1756,  the  very  same  year  that  M.  le  Sage  communicated  to  the 
Academy  of  Sciences  the  observation  on  the  subject  of  solid  angles,  men- 
tioned in  a  former  note  ;  and  it  is  singular,  that  these  two  Geometers,  with- 
out any  communication  with  one  another,  should  almost  at  the  same  time 
have  made  two  discoveries  very  nearly  connected,  yet  neither  of  them  com- 
prehending the  whole  truth,  so  that  each  is  imperfect  without  the  other. 

Dr.  Simson  has  shewn  the  truth  of  his  remark,  by  the  following  reason- 
ing. 

"  Let  there  be  any  plane  rectilineal  figure,  as  the  triangle  ABC,  and  from 
a  point  D  within  it,  draw  the  straight  line  DE  at  right  angles  to  the  plane 
ABC  ;  in  DE  take  DE,  DF  equal  to  one  another,  upon  the  opposite  sides 
of  the  plane,  and  let  G  be  any  point  in  EF  ;  join  DA ,  DB,  DC  ;  EA,  EB, 
EC  ;  FA,  FB,  FC  ;  GA,  GB,  GC  :  Because  the  straight  line  EDF  is  at 
right  angles  to  the  plane  ABC,  it  makes  right  angles  with  DA,  DB,  DC, 
which  it  meets  in  that  plane  ;  and  in  the  triangles  EDB,  FDB,  ED  and 
DB  are  equal  to  FD,  and  DB,  each  to  each,  and  they  contain  right  angles  ; 
therefore  the  base  EB  is  equal  to  the  base  FB  ;  in  the  same  manner  EA  is 


NOTES,  SUPPL.  BOOK  III.  315 


equal  to  FA,  and  EC  to  FC  :  and  in  the  triangles  EBA,  FBA,  EB,  BA  are 
equal  to  FB,  BA,  and  the  base  EA  is  equal  to  the  base  FA  ;  Avherefore 
the  angle  EBA  is  equal  to  the  angle  FBA,  and  the  triangle  EBA  equal 
to  the  triangle  FBA,  and  the  other  angles  equal  to  the  other  angles  ;  there- 
fore these  triangles  are  similar :  In  the  same  manner  the  triangle  EBC  is 
similar  to  the  triangle  FBC,  and  the  triangle  EAC  to  FAC  ;  therefore  there 
are  two  solid  figures,  each  of  which  is  contained  by  six  triangles,  one  of  them 
by  three  triangles,  the  common  vertex  of  which  is  the  point  G,  and  their 
bases  the  straight  lines  AB,  BC,  C  A,  and  by  three  other  triangles  the  com- 
mon vertex  of  which  is  the  point  E,  and  their  bases  the  same  lines  AB,  BC, 
CA.  The  other  solid  is  contained  by  the  same  three  triangles,  the  common 
vertex  of  which  is  G,  and  their  bases  AB,  BC,  CA  ;  and  by  three  other  tri- 
angles, of  which  the  common  vertex  is  the  point  F,  and  their  bases  the  same 
straight  lines  AB,  BC,  CA:  Now,  the  three  triangles  GAB,  GBC,  GCA 
are  common  to  both  solids,  and  the  three  others  EAB,  EBC,  ECA,  of  the 
first  solid  have  been  shown  to  be  equal  and  similar  to  the  three  others, 
FAB,  FBC,  FCA  of  the  other  solid,  each  to  each ;  therefore,  these  two 
solids  are  contained  by  the  same  number  of  equal  and  similar  planes  :  But 
that  they  are  not  equal  is  manifest,  because  the  first  of  them  is  contained  in 
the  other  :  Therefore  it  is  not  universally  true,  that  solids  are  equal  which 
are  contained  by  the  same  number  of  equal  and  similar  planes." 

"  CoR.  From  this  it  appears,  that  two  unequal  solid  angles  may  be  con- 
tained by  the  same  number  of  equal  plane  angles." 

"  For  the  solid  angle  at  B,  which  is  contained  by  the  four  plane  angles 
EBA,  EBC,  GBA,  GBC  is  not  equal  to  the  solid  angle  at  the  same  point 
B,  which  is  contained  by  the  four  plane  angles  FBA,  FBC,  GBA,  GBC  ; 
for  the  last  contains  the  other.  And  each  of  them  is  contained  by  four 
plane  angles,  which  are  equal  to  one  another,  each  to  each,  or  are  the  self- 
same, as  has  been  proved  :  And  indeed,  there  may  be  innumerable  solid 
angles  all  unequal  to  one  another,  which  are  each  of  them  contained  by 
plane  angles  that  are  equal  to  one  another,  each  to  each.  It  is  likewise 
manifest^  that  the  before-mentioned  solids  are  not  similar,  since  their  solid 
angles  are  not  all  equal." 


PLANE  TRIGONOMETRY. 


DEFINITIONS,  &c. 

Trigonometry  is  defined  in  tlie  text  to  be  the  application  of  Number 
to  express  tbe  relations  of  the  sides  and  angles  of  triangles.  It  depends 
therefore,  on  the  47th  of  the  first  of  Euclid,  and  on  the  7th  of  the  first  of  the 
Supplement,  the  two  propositions  which  do  most  immediately  connect 
together  the  sciences  of  Arithmetic  and  Geometry. 

The  sine  of  an  angle  is  defined  above  in  the  usual  way,  viz.  the  perpen- 
dicular drawn  from  one  extremity  of  the  arc,  which  measures  the  angle  on 
the  radius  passing  through  the  other  ;  but  in  strictness  the  sine  is  not  the 
perpendicular  itself,  but,  the  ratio  of  that  perpendicular  to  the  radius,  for  it 
is  this  ratio  which  remains  constant,  while  the  angle  continues  the  same, 
though  the  radius  vary.  It  might  be  convenient,  therefore,  to  define  the 
sine  to  be  the  quotient  which  arises  from  dividing  the  perpendicular  just 
described  by  the  radius  of  the  circle. 

So  also,  if  one  of  the  sides  of  a  right  angled  triangle  about  the  right  an- 
gle be  divided  by  the  other,  the  quotient  is  the  tangent  of  the  angle  op- 
posite to  the  first-mentioned  side,  &c.  But  though  this  is  certainly  the 
rigorous  way  of  conceiving  the  sines,  tangents,  &c.  of  angles,  which  are 
in  reality  not  magnitudes,  but  the  ratios  of  magnitudes  ;  yet  as  this  idea  is 
a  little  more  abstract  than  the  common  one,  and  would  also  involve  some 
change  in  the  language  of  Trigonometry,  at  the  same  time  that  it  would 
in  the  end  lead  to  nothing  that  is  not  attained  by  making  the  radius  equal 
to  unity,  I  have  adhered  to  the  common  method,  though  I  have  thought 
it  right  to  point  out  that  which  should  in  strictness  be  pursued. 

A  proposition  is  left  out  in  the  Plane  Trigonometry,  which  the  astro- 
nomers make  use  of  in  order,  when  two  sides  of  a  triangle,  and  the  angle 
contained  by  them,  are  given,  to  find  the  angles  at  the  base,  without 
making  use  of  the  sum  or  difference  of  the  sides,  which,  in  some  cases, 
when  only  the  Logarithms  of  the  sides  are  given,  cannot  be  conveniently 
found. 


,      NOTES.     PL.  TRIGONOMETRY.  317 

THEOREM. 

Iji  as  the  greater  of  any  two  sides  of  a  triangle  to  the  less,  so  the  radius  to  the 
tangent  of  a  certain  angle;  then  will  the  radius  he  to  the  tangent  of  the  diffe- 
rence between  that  angle  and  half  a  right  angle,  as  the  tangent  of  half  the 
sum  of  the  angles,  at  the  base  of  the  triangle  to  the  tangent  of  half  their 
difference. 

Let  ABC  be  a  triangle,  the  sides  of 
which  are  BC  and  CA,  and  the  base 
AB,  and  let  BC  be  greater  than  CA. 
Let  DC  be  drawn  at  right  angles  to 
BC,  and  equal  to  KG ;  join  BD,  and 
because  (Prop.  1.)  in  the  right  angled 
triangle  BCD,  BC  :  CD  :  :  R  :  tan 
CBD,  CBD  is  the  angle  of  which  the 
tangent  is  to  the  radius  as  CD  to  BC, 
that  is,  as  CA  to  BC,  or  as  the  least 
of  the  two  sides  of  the  triangle  to  the 
greatest. 

ButBC+CD  :  BC-CD  : :  tan  i(CDB  +  CBD)  : 
tan  i(CDB-CBD)  (Prop.  5.) ; 

and  also,  BC+CA  :  BC— CA  : :  tan  J  (CAB+CBA)  : 
tan  I  (CAB— CBA).     Therefore,  since  CD=CA, 
tan  I  (CDB+CBD)  :  tan  ^  (CDB— CBD) : : 
tan  \  (CAB  +  CBA)  :  tan  J  (CAB— CBA).     But  because  the 
angles  CDB4-CBD=90o,  tan  i(CDB4-CBD)  : 
tan  \  (CDB— CBD)  : :  R  :  tan  (45°- CBD),  (2  Cor.  Prop.  3.) , 
therefore,  R  :  tan  (45^- CBD)  :  :  tan  ^  (CAB+CBA)  : 
tan  J  (CAB — CBA) ;  and  CBD  was  already  shewn  to  be  such  an  angle 
that  BC  :  CA  : :  R  :  tan  CBD. 

CoR.  If  BC,  CA,  and  the  angle  0  arie  given  to  find  the  angles  A  and  B  ; 
find  an  angle  E  such,  that  BC  :  CA  : :  R  :  tan  E  ;  then  R  :  tan  (45° — E) 
: :  tan  \  (A+B)  :  tan  J  (A— B).  Thus  \  (A— B)  is  found,  and  J  (A+B) 
being  given,  A  and  B  are  each  of  them  known.     Lem.  2. 

In  reading  the  elements  of  Plane  Trigonometry,  it  may  be  of  use  to  ob- 
serve, that  the  first  five  propositions  contain  all  the  rules  absolutely  neces- 
sary for  solving  the  diflferent  cases  of  plane  triangles.  The  learner,  when 
he  studies  Trigonometry  for  the  first  time,  may  satisfy  himself  with  these 
propositions,  but  should  by  no  means  neglect  the  others  in  a  subsequen* 
perusal. 

PROP.  VII.  and  VIII. 

I  have  changed  the  demonstration  which  I  gave  of  these  propositions  in 
the  first  edition,  for  two  others  considerably  simpler  and  more  concise,  given 
me  by  Mr.  Jardine,  teacher  of  the  Mathematics  in  Edinburgh,  formerly 
one  of  my  pupils,  to  whose  ingenuity  and  skill  I  am  very  glad  to  bear  this 
public  testimony. 


SPHERICAL 
TRIGONOMETRY- 


PROP.  V. 

The  angles  at  the  base  of  an  isosceles  spherical  triangle  are  symmetrical 
magnitudes,  not  admitting  of  being  laid  on  one  another,  nor  of  coinciding, 
notwithstanding  their  equality.  It  might  be  considered  as  a  sufficient 
proof  that  they  are  equal,  to  observe  that  they  are  each  determined  to  be 
of  a  certain  magnitude  rather  than  any  other,  by  conditions  which  are  pre- 
cisely the  same,  so  that  there  is  no  reason  why  one  of  them  should  be 
greater  than  another.  For  the  sake  of  those  to  whom  this  reasoning  may 
not  prove  satisfactory,  the  demonstration  in  the  text  is  given,  which  is 
strictly  geometrical. 

For  the  demonstrations  of  the  two  propositions  that  are  given  in  the  end 
of  the  Appendix  to  the  Spherical  Trigonometry,  see  Elementa  Sphaericorum, 
Theor.  66.  apud  Wolfii  Opera  Math.  tom.  iii. ;  Trigonometrie  par  Cagnoli 
^  463  ;  Trigonometrie  Spherique  par  Mauduit,  ^  165. 


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